One of those things beginners in C find difficult is the concept of pointers. The purpose of this tutorial is to provide an introduction to pointers and their use to these beginners.
I have found that often the main reason
beginners have a problem with pointers is that they have a weak or minimal
feeling for variables, (as they are used in C). Thus we start with a discussion
of C variables in general.
A variable in a program is something with a
name, the value of which can vary. The way the compiler and linker handles this
is that it assigns a specific block of memory within the computer to hold the
value of that variable. The size of that block depends on the range over which
the variable is allowed to vary. For example, on PC's the size of an integer
variable is 2 bytes, and that of a long integer is 4 bytes. In C the size of a
variable type such as an integer need not be the same on all types of machines.
When we declare a variable we inform the
compiler of two things, the name of the variable and the type of the variable.
For example, we declare a variable of type integer with the name k by writing:
int k;
On seeing the "int" part of this
statement the compiler sets aside 2 bytes of memory (on a PC) to hold the value
of the integer. It also sets up a symbol table. In that table it adds the
symbol k and the relative address in memory where those 2 bytes were set aside.
Thus, later if we write:
k = 2;
we expect that, at run time when this
statement is executed, the value 2 will be placed in that memory location reserved
for the storage of the value of k. In C we refer to a variable such as the
integer k as an "object".
In a sense there are two "values"
associated with the object k. One is the value of the integer stored there (2
in the above example) and the other the "value" of the memory
location, i.e., the address of k. Some texts refer to these two values with the
nomenclature rvalue (right value, pronounced "are value") and lvalue
(left value, pronounced "el value") respectively.
In some languages, the lvalue is the value
permitted on the left side of the assignment operator '=' (i.e. the address
where the result of evaluation of the right side ends up). The rvalue is that
which is on the right side of the assignment statement, the 2 above. Rvalues
cannot be used on the left side of the assignment statement. Thus: 2 = k; is
illegal.
Actually, the above definition of
"lvalue" is somewhat modified for C. According to K&R II (page
197): [1]
"An object is a named region of storage;
an lvalue is an expression referring to an object."
However, at this point, the definition
originally cited above is sufficient. As we become more familiar with pointers
we will go into more detail on this.
Okay, now consider:
int j, k;
k = 2;
j = 7; <-- line 1
k = j; <-- line 2
In the above, the compiler interprets the j
in line 1 as the address of the variable j (its lvalue) and creates code to
copy the value 7 to that address. In line 2, however, the j is interpreted as
its rvalue (since it is on the right hand side of the assignment operator '=').
That is, here the j refers to the value stored at the memory location set aside
for j, in this case 7. So, the 7 is copied to the address designated by the
lvalue of k.
In all of these examples, we are using 2 byte
integers so all copying of rvalues from one storage location to the other is
done by copying 2 bytes. Had we been using long integers, we would be copying 4
bytes.
Now, let's say that we have a reason for
wanting a variable designed to hold an lvalue (an address). The size required
to hold such a value depends on the system. On older desk top computers with
64K of memory total, the address of any point in memory can be contained in 2
bytes. Computers with more memory would require more bytes to hold an address.
Some computers, such as the IBM PC might require special handling to hold a
segment and offset under certain circumstances. The actual size required is not
too important so long as we have a way of informing the compiler that what we
want to store is an address.
Such a variable is called a pointer variable
(for reasons which hopefully will become clearer a little later). In C when we
define a pointer variable we do so by preceding its name with an asterisk. In C
we also give our pointer a type, which, in this case, refers to the type of
data stored at the address we will be storing in our pointer. For example,
consider the variable declaration:
int *ptr;
ptr is the name of our variable (just as k
was the name of our integer variable). The '*' informs the compiler that we
want a pointer variable, i.e. to set aside however many bytes is required to
store an address in memory. The int says that we intend to use our pointer
variable to store the address of an integer. Such a pointer is said to "point
to" an integer. However, note that when we wrote int k; we did not give k
a value. If this definition is made outside of any function ANSI compliant
compilers will initialize it to zero. Similarly, ptr has no value, that is we
haven't stored an address in it in the above declaration. In this case, again
if the declaration is outside of any function, it is initialized to a value
guaranteed in such a way that it is guaranteed to not point to any C object or
function. A pointer initialized in this manner is called a "null"
pointer.
The actual bit pattern used for a null
pointer may or may not evaluate to zero since it depends on the specific system
on which the code is developed. To make the source code compatible between
various compilers on various systems, a macro is used to represent a null
pointer. That macro goes under the name NULL. Thus, setting the value of a
pointer using the NULL macro, as with an assignment statement such as ptr =
NULL, guarantees that the pointer has become a null pointer. Similarly, just as
one can test for an integer value of zero, as in if(k == 0), we can test for a
null pointer using if (ptr == NULL).
But, back to using our new variable ptr.
Suppose now that we want to store in ptr the address of our integer variable k.
To do this we use the unary & operator and write:
ptr = &k;
What the & operator does is retrieve the
lvalue (address) of k, even though k is on the right hand side of the
assignment operator '=', and copies that to the contents of our pointer ptr.
Now, ptr is said to "point to" k. Bear with us now, there is only one
more operator we need to discuss.
The "dereferencing operator" is the
asterisk and it is used as follows:
*ptr = 7;
will copy 7 to the address pointed to by ptr.
Thus if ptr "points to" (contains the address of) k, the above
statement will set the value of k to 7. That is, when we use the '*' this way
we are referring to the value of that which ptr is pointing to, not the value
of the pointer itself.
Similarly, we could write:
printf("%d\n",*ptr);
to print to the screen the integer value
stored at the address pointed to by ptr;.
One way to see how all this stuff fits
together would be to run the following program and then review the code and the
output carefully.
------------ Program 1.1 ---------------------------------
/* Program 1.1 from PTRTUT10.TXT 6/10/97 */
#include <stdio.h>
int j, k;
int *ptr;
int main(void)
{
j = 1;
k = 2;
ptr = &k;
printf("\n");
printf("j has the value %d and is stored at %p\n", j, (void *)&j);
printf("k has the value %d and is stored at %p\n", k, (void *)&k);
printf("ptr has the value %p and is stored at %p\n", ptr, (void *)&ptr);
printf("The value of the integer pointed to by ptr is %d\n", *ptr);
return 0;
}
Note: We have yet to discuss those aspects of
C which require the use of the (void *) expression used here. For now, include
it in your test code. We'll explain the reason behind this expression later.
To review:
- A
variable is declared by giving it a type and a name (e.g. int k;)
- A
pointer variable is declared by giving it a type and a name (e.g. int
*ptr) where the asterisk tells the compiler that the variable named ptr is
a pointer variable and the type tells the compiler what type the pointer
is to point to (integer in this case).
- Once a
variable is declared, we can get its address by preceding its name with
the unary & operator, as in &k.
- We can
"dereference" a pointer, i.e. refer to the value of that which
it points to, by using the unary '*' operator as in *ptr.
- An
"lvalue" of a variable is the value of its address, i.e. where
it is stored in memory. The "rvalue" of a variable is the value
stored in that variable (at that address).
CHAPTER 2:
Pointer types and Arrays
Okay, let's move on. Let us consider why we
need to identify the type of variable that a pointer points to, as in:
int *ptr;
One reason for doing this is so that later,
once ptr "points to" something, if we write:
*ptr = 2;
the compiler will know how many bytes to copy
into that memory location pointed to by ptr. If ptr was declared as pointing to
an integer, 2 bytes would be copied, if a long, 4 bytes would be copied.
Similarly for floats and doubles the appropriate number will be copied. But,
defining the type that the pointer points to permits a number of other
interesting ways a compiler can interpret code. For example, consider a block
in memory consisting if ten integers in a row. That is, 20 bytes of memory are
set aside to hold 10 integers.
Now, let's say we point our integer pointer
ptr at the first of these integers. Furthermore lets say that integer is
located at memory location 100 (decimal). What happens when we write:
ptr + 1;
Because the compiler "knows" this
is a pointer (i.e. its value is an address) and that it points to an integer
(its current address, 100, is the address of an integer), it adds 2 to ptr
instead of 1, so the pointer "points to" the next integer, at memory
location 102. Similarly, were the ptr declared as a pointer to a long, it would
add 4 to it instead of 1. The same goes for other data types such as floats,
doubles, or even user defined data types such as structures. This is obviously
not the same kind of "addition" that we normally think of. In C it is
referred to as addition using "pointer arithmetic", a term which we
will come back to later.
Similarly, since ++ptr and ptr++ are both
equivalent to ptr + 1 (though the point in the program when ptr is incremented
may be different), incrementing a pointer using the unary ++ operator, either
pre- or post-, increments the address it stores by the amount sizeof(type)
where "type" is the type of the object pointed to. (i.e. 2 for an
integer, 4 for a long, etc.).
Since a block of 10 integers located
contiguously in memory is, by definition, an array of integers, this brings up
an interesting relationship between arrays and pointers.
Consider the following:
int my_array[] = {1,23,17,4,-5,100};
Here we have an array containing 6 integers.
We refer to each of these integers by means of a subscript to my_array, i.e.
using my_array[0] through my_array[5]. But, we could alternatively access them
via a pointer as follows:
int *ptr;
ptr = &my_array[0]; /* point our pointer at the first
integer in our array */
And then we could print out our array either
using the array notation or by dereferencing our pointer. The following code
illustrates this:
----------- Program 2.1 -----------------------------------
/* Program 2.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int my_array[] = {1,23,17,4,-5,100};
int *ptr;
int main(void)
{
int i;
ptr = &my_array[0]; /* point our pointer to the first
element of the array */
printf("\n\n");
for (i = 0; i < 6; i++)
{
printf("my_array[%d] = %d ",i,my_array[i]); /*<-- A */
printf("ptr + %d = %d\n",i, *(ptr + i)); /*<-- B */
}
return 0;
}
Compile and run the above program and
carefully note lines A and B and that the program prints out the same values in
either case. Also observe how we dereferenced our pointer in line B, i.e. we
first added i to it and then dereferenced the new pointer. Change line B to
read:
printf("ptr + %d = %d\n",i, *ptr++);
and run it again... then change it to:
printf("ptr + %d = %d\n",i, *(++ptr));
and try once more. Each time try and predict
the outcome and carefully look at the actual outcome.
In C, the standard states that wherever we
might use &var_name[0] we can replace that with var_name, thus in our code
where we wrote:
ptr = &my_array[0];
we can write:
ptr = my_array;
to achieve the same result.
This leads many texts to state that the name
of an array is a pointer. I prefer to mentally think "the name of the
array is the address of first element in the array". Many beginners
(including myself when I was learning) have a tendency to become confused by
thinking of it as a pointer. For example, while we can write
ptr = my_array;
we cannot write
my_array = ptr;
The reason is that while ptr is a variable,
my_array is a constant. That is, the location at which the first element of
my_array will be stored cannot be changed once my_array[] has been declared.
Earlier when discussing the term
"lvalue" I cited K&R-2 where it stated:
"An object is a named region of storage;
an lvalue is an expression referring to an object".
This raises an interesting problem. Since
my_array is a named region of storage, why is my_array in the above assignment
statement not an lvalue? To resolve this problem, some refer to my_array as an
"unmodifiable lvalue".
Modify the example program above by changing
ptr = &my_array[0];
to
ptr = my_array;
and run it again to verify the results are
identical.
Now, let's delve a little further into the
difference between the names ptr and my_array as used above. Some writers will
refer to an array's name as a constant pointer. What do we mean by that? Well,
to understand the term "constant" in this sense, let's go back to our
definition of the term "variable". When we declare a variable we set
aside a spot in memory to hold the value of the appropriate type. Once that is
done the name of the variable can be interpreted in one of two ways. When used
on the left side of the assignment operator, the compiler interprets it as the
memory location to which to move that value resulting from evaluation of the
right side of the assignment operator. But, when used on the right side of the
assignment operator, the name of a variable is interpreted to mean the contents
stored at that memory address set aside to hold the value of that variable.
With that in mind, let's now consider the
simplest of constants, as in:
int i, k;
i = 2;
Here, while i is a variable and then occupies
space in the data portion of memory, 2 is a constant and, as such, instead of
setting aside memory in the data segment, it is imbedded directly in the code
segment of memory. That is, while writing something like k = i; tells the
compiler to create code which at run time will look at memory location &i
to determine the value to be moved to k, code created by i = 2; simply puts the
2 in the code and there is no referencing of the data segment. That is, both k
and i are objects, but 2 is not an object.
Similarly, in the above, since my_array is a
constant, once the compiler establishes where the array itself is to be stored,
it "knows" the address of my_array[0] and on seeing:
ptr = my_array;
it simply uses this address as a constant in
the code segment and there is no referencing of the data segment beyond that.
This might be a good place explain further
the use of the (void *) expression used in Program 1.1 of Chapter 1. As we have
seen we can have pointers of various types. So far we have discussed pointers
to integers and pointers to characters. In coming chapters we will be learning
about pointers to structures and even pointer to pointers.
Also we have learned that on different
systems the size of a pointer can vary. As it turns out it is also possible
that the size of a pointer can vary depending on the data type of the object to
which it points. Thus, as with integers where you can run into trouble
attempting to assign a long integer to a variable of type short integer, you
can run into trouble attempting to assign the values of pointers of various
types to pointer variables of other types.
To minimize this problem, C provides for a
pointer of type void. We can declare such a pointer by writing:
void *vptr;
A void pointer is sort of a generic pointer.
For example, while C will not permit the comparison of a pointer to type
integer with a pointer to type character, for example, either of these can be
compared to a void pointer. Of course, as with other variables, casts can be
used to convert from one type of pointer to another under the proper
circumstances. In Program 1.1. of Chapter 1 I cast the pointers to integers
into void pointers to make them compatible with the %p conversion
specification. In later chapters other casts will be made for reasons defined
therein.
Well, that's a lot of technical stuff to
digest and I don't expect a beginner to understand all of it on first reading.
With time and experimentation you will want to come back and re-read the first
2 chapters. But for now, let's move on to the relationship between pointers,
character arrays, and strings.
CHAPTER 3:
Pointers and Strings
The study of strings is useful to further tie
in the relationship between pointers and arrays. It also makes it easy to
illustrate how some of the standard C string functions can be implemented.
Finally it illustrates how and when pointers can and should be passed to
functions.
In C, strings are arrays of characters. This
is not necessarily true in other languages. In BASIC, Pascal, Fortran and
various other languages, a string has its own data type. But in C it does not.
In C a string is an array of characters terminated with a binary zero character
(written as '\0'). To start off our discussion we will write some code which,
while preferred for illustrative purposes, you would probably never write in an
actual program. Consider, for example:
char my_string[40];
my_string[0] = 'T';
my_string[1] = 'e';
my_string[2] = 'd':
my_string[3] = '\0';
While one would never build a string like
this, the end result is a string in that it is an array of characters
terminated with a nul character. By definition, in C, a string is an array of
characters terminated with the nul character. Be aware that "nul" is
not the same as "NULL". The nul refers to a zero as defined by the
escape sequence '\0'. That is it occupies one byte of memory. NULL, on the
other hand, is the name of the macro used to initialize null pointers. NULL is
#defined in a header file in your C compiler, nul may not be #defined at all.
Since writing the above code would be very
time consuming, C permits two alternate ways of achieving the same thing.
First, one might write:
char my_string[40] = {'T', 'e', 'd', '\0',};
But this also takes more typing than is
convenient. So, C permits:
char my_string[40] = "Ted";
When the double quotes are used, instead of
the single quotes as was done in the previous examples, the nul character (
'\0' ) is automatically appended to the end of the string.
In all of the above cases, the same thing
happens. The compiler sets aside an contiguous block of memory 40 bytes long to
hold characters and initialized it such that the first 4 characters are Ted\0.
Now, consider the following program:
------------------Program 3.1-------------------------------------
/* Program 3.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
char strA[80] = "A string to be used for demonstration purposes";
char strB[80];
int main(void)
{
char *pA; /* a pointer to type character */
char *pB; /* another pointer to type character */
puts(strA); /* show string A */
pA = strA; /* point pA at string A */
puts(pA); /* show what pA is pointing to */
pB = strB; /* point pB at string B */
putchar('\n'); /* move down one line on the screen */
while(*pA != '\0') /* line A (see text) */
{
*pB++ = *pA++; /* line B (see text) */
}
*pB = '\0'; /* line C (see text) */
puts(strB); /* show strB on screen */
return 0;
}
--------- end program 3.1 -------------------------------------
In the above we start out by defining two
character arrays of 80 characters each. Since these are globally defined, they
are initialized to all '\0's first. Then, strA has the first 42 characters
initialized to the string in quotes.
Now, moving into the code, we declare two
character pointers and show the string on the screen. We then "point"
the pointer pA at strA. That is, by means of the assignment statement we copy
the address of strA[0] into our variable pA. We now use puts() to show that
which is pointed to by pA on the screen. Consider here that the function
prototype for puts() is:
int puts(const char *s);
For the moment, ignore the const. The
parameter passed to puts() is a pointer, that is the value of a pointer (since
all parameters in C are passed by value), and the value of a pointer is the
address to which it points, or, simply, an address. Thus when we write
puts(strA); as we have seen, we are passing the address of strA[0].
Similarly, when we write puts(pA); we are
passing the same address, since we have set pA = strA;
Given that, follow the code down to the
while() statement on line A. Line A states:
While the character pointed to by pA (i.e.
*pA) is not a nul character (i.e. the terminating '\0'), do the following:
Line B states: copy the character pointed to
by pA to the space pointed to by pB, then increment pA so it points to the next
character and pB so it points to the next space.
When we have copied the last character, pA
now points to the terminating nul character and the loop ends. However, we have
not copied the nul character. And, by definition a string in C must be nul
terminated. So, we add the nul character with line C.
It is very educational to run this program
with your debugger while watching strA, strB, pA and pB and single stepping
through the program. It is even more educational if instead of simply defining
strB[] as has been done above, initialize it also with something like:
strB[80] = "12345678901234567890123456789012345678901234567890"
where the number of digits used is greater
than the length of strA and then repeat the single stepping procedure while
watching the above variables. Give these things a try!
Getting back to the prototype for puts() for
a moment, the "const" used as a parameter modifier informs the user
that the function will not modify the string pointed to by s, i.e. it will
treat that string as a constant.
Of course, what the above program illustrates
is a simple way of copying a string. After playing with the above until you
have a good understanding of what is happening, we can proceed to creating our
own replacement for the standard strcpy() that comes with C. It might look
like:
char *my_strcpy(char *destination, char *source)
{
char *p = destination;
while (*source != '\0')
{
*p++ = *source++;
}
*p = '\0';
return destination;
}
In this case, I have followed the practice
used in the standard routine of returning a pointer to the destination.
Again, the function is designed to accept the
values of two character pointers, i.e. addresses, and thus in the previous
program we could write:
int main(void)
{
my_strcpy(strB, strA);
puts(strB);
}
I have deviated slightly from the form used in standard C which would have
the prototype:
char *my_strcpy(char *destination, const char *source);
Here the "const" modifier is used
to assure the user that the function will not modify the contents pointed to by
the source pointer. You can prove this by modifying the function above, and its
prototype, to include the "const" modifier as shown. Then, within the
function you can add a statement which attempts to change the contents of that
which is pointed to by source, such as:
*source = 'X';
which would normally change the first
character of the string to an X. The const modifier should cause your compiler
to catch this as an error. Try it and see.
Now, let's consider some of the things the
above examples have shown us. First off, consider the fact that *ptr++ is to be
interpreted as returning the value pointed to by ptr and then incrementing the
pointer value. This has to do with the precedence of the operators. Were we to
write (*ptr)++ we would increment, not the pointer, but that which the pointer
points to! i.e. if used on the first character of the above example string the
'T' would be incremented to a 'U'. You can write some simple example code to
illustrate this.
Recall again that a string is nothing more
than an array of characters, with the last character being a '\0'. What we have
done above is deal with copying an array. It happens to be an array of
characters but the technique could be applied to an array of integers, doubles,
etc. In those cases, however, we would not be dealing with strings and hence
the end of the array would not be marked with a special value like the nul
character. We could implement a version that relied on a special value to
identify the end. For example, we could copy an array of positive integers by
marking the end with a negative integer. On the other hand, it is more usual
that when we write a function to copy an array of items other than strings we
pass the function the number of items to be copied as well as the address of
the array, e.g. something like the following prototype might indicate:
void int_copy(int *ptrA, int *ptrB, int nbr);
where nbr is the number of integers to be
copied. You might want to play with this idea and create an array of integers
and see if you can write the function int_copy() and make it work.
This permits using functions to manipulate
large arrays. For example, if we have an array of 5000 integers that we want to
manipulate with a function, we need only pass to that function the address of
the array (and any auxiliary information such as nbr above, depending on what
we are doing). The array itself does not get passed, i.e. the whole array is
not copied and put on the stack before calling the function, only its address
is sent.
This is different from passing, say an
integer, to a function. When we pass an integer we make a copy of the integer,
i.e. get its value and put it on the stack. Within the function any
manipulation of the value passed can in no way effect the original integer. But,
with arrays and pointers we can pass the address of the variable and hence
manipulate the values of the original variables.
CHAPTER 4:
More on Strings
Well, we have progressed quite a way
in a short time! Let's back up a little and look at what was done in Chapter 3
on copying of strings but in a different light. Consider the following
function:
char *my_strcpy(char dest[], char source[])
{
int i = 0;
while (source[i] != '\0')
{
dest[i] = source[i];
i++;
}
dest[i] = '\0';
return dest;
}
Recall that strings are arrays of
characters. Here we have chosen to use array notation instead of pointer
notation to do the actual copying. The results are the same, i.e. the string gets
copied using this notation just as accurately as it did before. This raises
some interesting points which we will discuss.
Since parameters are passed by value, in both
the passing of a character pointer or the name of the array as above, what
actually gets passed is the address of the first element of each array. Thus,
the numerical value of the parameter passed is the same whether we use a
character pointer or an array name as a parameter. This would tend to imply
that somehow source[i] is the same as *(p+i).
In fact, this is true, i.e wherever one
writes a[i] it can be replaced with *(a + i) without any problems. In fact, the
compiler will create the same code in either case. Thus we see that pointer
arithmetic is the same thing as array indexing. Either syntax produces the same
result.
This is NOT saying that pointers and arrays
are the same thing, they are not. We are only saying that to identify a given
element of an array we have the choice of two syntaxes, one using array
indexing and the other using pointer arithmetic, which yield identical results.
Now, looking at this last expression, part of
it.. (a + i), is a simple addition using the + operator and the rules of C
state that such an expression is commutative. That is (a + i) is identical to
(i + a). Thus we could write *(i + a) just as easily as *(a + i).
But *(i + a) could have come from i[a] ! From
all of this comes the curious truth that if:
char a[20];
int i;
writing
a[3] = 'x';
is the same as writing
3[a] = 'x';
Try it! Set up an array of characters,
integers or longs, etc. and assigned the 3rd or 4th element a value using the
conventional approach and then print out that value to be sure you have that
working. Then reverse the array notation as I have done above. A good compiler
will not balk and the results will be identical. A curiosity... nothing more!
Now, looking at our function above, when we
write:
dest[i] = source[i];
due to the fact that array indexing
and pointer arithmetic yield identical results, we can write this as:
*(dest + i) = *(source + i);
But, this takes 2 additions for each
value taken on by i. Additions, generally speaking, take more time than
incrementations (such as those done using the ++ operator as in i++). This may
not be true in modern optimizing compilers, but one can never be sure. Thus,
the pointer version may be a bit faster than the array version.
Another way to speed up the pointer version
would be to change:
while (*source != '\0')
to simply
while (*source)
since the value within the parenthesis
will go to zero (FALSE) at the same time in either case.
At this point you might want to experiment a
bit with writing some of your own programs using pointers. Manipulating strings
is a good place to experiment. You might want to write your own versions of
such standard functions as:
strlen();
strcat();
strchr();
and any others you might have on your
system.
We will come back to strings and their
manipulation through pointers in a future chapter. For now, let's move on and
discuss structures for a bit.
CHAPTER 5:
Pointers and Structures
As
you may know, we can declare the form of a block of data containing different
data types by means of a structure declaration. For example, a personnel file
might contain structures which look something like: struct tag {
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
Let's say we have a bunch of these structures
in a disk file and we want to read each one out and print out the first and
last name of each one so that we can have a list of the people in our files.
The remaining information will not be printed out. We will want to do this
printing with a function call and pass to that function a pointer to the
structure at hand. For demonstration purposes I will use only one structure for
now. But realize the goal is the writing of the function, not the reading of
the file which, presumably, we know how to do.
For review, recall that we can access
structure members with the dot operator as in:
--------------- program 5.1 ------------------
/* Program 5.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
struct tag {
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
struct tag my_struct; /* declare the structure my_struct */
int main(void)
{
strcpy(my_struct.lname,"Jensen");
strcpy(my_struct.fname,"Ted");
printf("\n%s ",my_struct.fname);
printf("%s\n",my_struct.lname);
return 0;
}
-------------- end of program 5.1 --------------
Now, this particular structure is rather
small compared to many used in C programs. To the above we might want to add:
date_of_hire; (data types not shown)
date_of_last_raise;
last_percent_increase;
emergency_phone;
medical_plan;
Social_S_Nbr;
etc.....
If we have a large number of employees, what
we want to do is manipulate the data in these structures by means of functions.
For example we might want a function print out the name of the employee listed
in any structure passed to it. However, in the original C (Kernighan &
Ritchie, 1st Edition) it was not possible to pass a structure, only a pointer
to a structure could be passed. In ANSI C, it is now permissible to pass the
complete structure. But, since our goal here is to learn more about pointers,
we won't pursue that.
Anyway, if we pass the whole structure it
means that we must copy the contents of the structure from the calling function
to the called function. In systems using stacks, this is done by pushing the
contents of the structure on the stack. With large structures this could prove
to be a problem. However, passing a pointer uses a minimum amount of stack
space.
In any case, since this is a discussion of
pointers, we will discuss how we go about passing a pointer to a structure and
then using it within the function.
Consider the case described, i.e. we want a
function that will accept as a parameter a pointer to a structure and from
within that function we want to access members of the structure. For example we
want to print out the name of the employee in our example structure.
Okay, so we know that our pointer is going to
point to a structure declared using struct tag. We declare such a pointer with
the declaration:
struct tag *st_ptr;
and we point it to our example structure
with:
st_ptr = &my_struct;
Now, we can access a given member by
de-referencing the pointer. But, how do we de-reference the pointer to a
structure? Well, consider the fact that we might want to use the pointer to set
the age of the employee. We would write:
(*st_ptr).age = 63;
Look at this carefully. It says, replace that
within the parenthesis with that which st_ptr points to, which is the structure
my_struct. Thus, this breaks down to the same as my_struct.age.
However, this is a fairly often used
expression and the designers of C have created an alternate syntax with the
same meaning which is:
st_ptr->age = 63;
With that in mind, look at the following
program:
------------ program 5.2 ---------------------
/* Program 5.2 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
struct tag{ /* the structure type */
char lname[20]; /* last name */
char fname[20]; /* first name */
int age; /* age */
float rate; /* e.g. 12.75 per hour */
};
struct tag my_struct; /* define the structure */
void show_name(struct tag *p); /* function prototype */
int main(void)
{
struct tag *st_ptr; /* a pointer to a structure */
st_ptr = &my_struct; /* point the pointer to my_struct */
strcpy(my_struct.lname,"Jensen");
strcpy(my_struct.fname,"Ted");
printf("\n%s ",my_struct.fname);
printf("%s\n",my_struct.lname);
my_struct.age = 63;
show_name(st_ptr); /* pass the pointer */
return 0;
}
void show_name(struct tag *p)
{
printf("\n%s ", p->fname); /* p points to a structure */
printf("%s ", p->lname);
printf("%d\n", p->age);
}
-------------------- end of program 5.2 ----------------
Again, this is a lot of information to absorb
at one time. The reader should compile and run the various code snippets and
using a debugger monitor things like my_struct and p while single stepping
through the main and following the code down into the function to see what is
happening.
CHAPTER 6:
Some more on Strings, and Arrays of Strings
Well, let's go back to strings for a bit. In
the following all assignments are to be understood as being global, i.e. made
outside of any function, including main().
We pointed out in an earlier chapter that we
could write:
char my_string[40] = "Ted";
which would allocate space for a 40 byte
array and put the string in the first 4 bytes (three for the characters in the
quotes and a 4th to handle the terminating '\0').
Actually, if all we wanted to do was store
the name "Ted" we could write:
char my_name[] = "Ted";
and the compiler would count the characters,
leave room for the nul character and store the total of the four characters in
memory the location of which would be returned by the array name, in this case
my_name.
In some code, instead of the above, you might
see:
char *my_name = "Ted";
which is an alternate approach. Is there a
difference between these? The answer is.. yes. Using the array notation 4 bytes
of storage in the static memory block are taken up, one for each character and
one for the terminating nul character. But, in the pointer notation the same 4
bytes required, plus N bytes to store the pointer variable my_name (where N
depends on the system but is usually a minimum of 2 bytes and can be 4 or
more).
In the array notation, my_name is short for
&myname[0] which is the address of the first element of the array. Since
the location of the array is fixed during run time, this is a constant (not a
variable). In the pointer notation my_name is a variable. As to which is the
better method, that depends on what you are going to do within the rest of the
program.
Let's now go one step further and consider
what happens if each of these declarations are done within a function as
opposed to globally outside the bounds of any function.
void my_function_A(char *ptr)
{
char a[] = "ABCDE"
.
.
}
void my_function_B(char *ptr)
{
char *cp = "FGHIJ"
.
.
}
In the case of my_function_A, the content, or
value(s), of the array a[] is considered to be the data. The array is said to
be initialized to the values ABCDE. In the case of my_function_B, the value of
the pointer cp is considered to be the data. The pointer has been initialized
to point to the string FGHIJ. In both my_function_A and my_function_B the
definitions are local variables and thus the string ABCDE is stored on the
stack, as is the value of the pointer cp. The string FGHIJ can be stored
anywhere. On my system it gets stored in the data segment.
By the way, array initialization of automatic
variables as I have done in my_function_A was illegal in the older K&R C
and only "came of age" in the newer ANSI C. A fact that may be
important when one is considering portability and backwards compatibility.
As long as we are discussing the
relationship/differences between pointers and arrays, let's move on to
multi-dimensional arrays. Consider, for example the array:
char multi[5][10];
Just what does this mean? Well, let's
consider it in the following light.
char multi[5][10];
Let's take the underlined part to be the
"name" of an array. Then prepending the char and appending the [10]
we have an array of 10 characters. But, the name multi[5] is itself an array
indicating that there are 5 elements each being an array of 10 characters.
Hence we have an array of 5 arrays of 10 characters each..
Assume we have filled this two dimensional
array with data of some kind. In memory, it might look as if it had been formed
by initializing 5 separate arrays using something like:
multi[0] = {'0','1','2','3','4','5','6','7','8','9'}
multi[1] = {'a','b','c','d','e','f','g','h','i','j'}
multi[2] = {'A','B','C','D','E','F','G','H','I','J'}
multi[3] = {'9','8','7','6','5','4','3','2','1','0'}
multi[4] = {'J','I','H','G','F','E','D','C','B','A'}
At the same time, individual elements might
be addressable using syntax such as:
multi[0][3] = '3'
multi[1][7] = 'h'
multi[4][0] = 'J'
Since arrays are contiguous in memory, our
actual memory block for the above should look like:
0123456789abcdefghijABCDEFGHIJ9876543210JIHGFEDCBA
^
|_____ starting at the address &multi[0][0]
Note that I did not write multi[0] =
"0123456789". Had I done so a terminating '\0' would have been
implied since whenever double quotes are used a '\0' character is appended to
the characters contained within those quotes. Had that been the case I would
have had to set aside room for 11 characters per row instead of 10.
My goal in the above is to illustrate how
memory is laid out for 2 dimensional arrays. That is, this is a 2 dimensional
array of characters, NOT an array of "strings".
Now, the compiler knows how many columns are
present in the array so it can interpret multi + 1 as the address of the 'a' in
the 2nd row above. That is, it adds 10, the number of columns, to get this
location. If we were dealing with integers and an array with the same dimension
the compiler would add 10*sizeof(int) which, on my machine, would be 20. Thus,
the address of the 9 in the 4th row above would be &multi[3][0] or *(multi
+ 3) in pointer notation. To get to the content of the 2nd element in the 4th
row we add 1 to this address and dereference the result as in
*(*(multi + 3) + 1)
With a little thought we can see that:
*(*(multi + row) + col) and
multi[row][col] yield the same results.
The following program illustrates this using
integer arrays instead of character arrays.
------------------- program 6.1 ----------------------
/* Program 6.1 from PTRTUT10.HTM 6/13/97*/
#include <stdio.h>
#define ROWS 5
#define COLS 10
int multi[ROWS][COLS];
int main(void)
{
int row, col;
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
multi[row][col] = row*col;
}
}
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
printf("\n%d ",multi[row][col]);
printf("%d ",*(*(multi + row) + col));
}
}
return 0;
}
----------------- end of program 6.1 ---------------------
Because of the double de-referencing required
in the pointer version, the name of a 2 dimensional array is often said to be
equivalent to a pointer to a pointer. With a three dimensional array we would
be dealing with an array of arrays of arrays and some might say its name would
be equivalent to a pointer to a pointer to a pointer. However, here we have
initially set aside the block of memory for the array by defining it using
array notation. Hence, we are dealing with a constant, not a variable. That is
we are talking about a fixed address not a variable pointer. The dereferencing
function used above permits us to access any element in the array of arrays
without the need of changing the value of that address (the address of
multi[0][0] as given by the symbol multi).
CHAPTER 7:
More on Multi-Dimensional Arrays
In the previous chapter we noted that
given
#define ROWS 5
#define COLS 10
int multi[ROWS][COLS];
we can access individual elements of
the array multi using either:
multi[row][col]
or
*(*(multi + row) + col)
To understand more fully what is going
on, let us replace
*(multi + row)
with X as in:
*(X + col)
Now, from this we see that X is like a
pointer since the expression is de-referenced and we know that col is an
integer. Here the arithmetic being used is of a special kind called
"pointer arithmetic" is being used. That means that, since we are
talking about an integer array, the address pointed to by (i.e. value of) X +
col + 1 must be greater than the address X + col by and amount equal to
sizeof(int).
Since we know the memory layout for 2
dimensional arrays, we can determine that in the expression multi + row as used
above, multi + row + 1 must increase by value an amount equal to that needed to
"point to" the next row, which in this case would be an amount equal
to COLS * sizeof(int).
That says that if the expression *(*(multi +
row) + col) is to be evaluated correctly at run time, the compiler must
generate code which takes into consideration the value of COLS, i.e. the 2nd
dimension. Because of the equivalence of the two forms of expression, this is
true whether we are using the pointer expression as here or the array
expression multi[row][col].
Thus, to evaluate either expression, a total
of 5 values must be known:
- The
address of the first element of the array, which is returned by the
expression multi, i.e., the name of the array.
- The
size of the type of the elements of the array, in this case sizeof(int).
- The 2nd
dimension of the array
- The
specific index value for the first dimension, row in this case.
- The
specific index value for the second dimension, col in this case.
Given all of that, consider the problem
of designing a function to manipulate the element values of a previously
declared array. For example, one which would set all the elements of the array
multi to the value 1.
void set_value(int m_array[][COLS])
{
int row, col;
for (row = 0; row < ROWS; row++)
{
for (col = 0; col < COLS; col++)
{
m_array[row][col] = 1;
}
}
}
And to call this function we would
then use:
set_value(multi);
Now, within the function we have used
the values #defined by ROWS and COLS that set the limits on the for loops. But,
these #defines are just constants as far as the compiler is concerned, i.e.
there is nothing to connect them to the array size within the function. row and
col are local variables, of course. The formal parameter definition permits the
compiler to determine the characteristics associated with the pointer value
that will be passed at run time. We really don’t need the first dimension and,
as will be seen later, there are occasions where we would prefer not to define
it within the parameter definition, out of habit or consistency, I have not
used it here. But, the second dimension must be used as has been shown in the
expression for the parameter. The reason is that we need this in the evaluation
of m_array[row][col] as has been described. While the parameter defines the
data type (int in this case) and the automatic variables for row and column are
defined in the for loops, only one value can be passed using a single
parameter. In this case, that is the value of multi as noted in the call
statement, i.e. the address of the first element, often referred to as a
pointer to the array. Thus, the only way we have of informing the compiler of
the 2nd dimension is by explicitly including it in the parameter definition.
In fact, in general all dimensions of higher
order than one are needed when dealing with multi-dimensional arrays. That is
if we are talking about 3 dimensional arrays, the 2nd and 3rd dimension must be
specified in the parameter definition.
CHAPTER 8:
Pointers to Arrays
Pointers, of course, can be
"pointed at" any type of data object, including arrays. While that
was evident when we discussed program 3.1, it is important to expand on how we
do this when it comes to multi-dimensional arrays.
To review, in Chapter 2 we stated that given
an array of integers we could point an integer pointer at that array using:
int *ptr;
ptr = &my_array[0]; /* point our pointer at the first
integer in our array */
As we stated there, the type of the
pointer variable must match the type of the first element of the array.
In addition, we can use a pointer as a formal
parameter of a function which is designed to manipulate an array. e.g.
Given:
int array[3] = {'1', '5', '7'};
void a_func(int *p);
Some programmers might prefer to write
the function prototype as:
void a_func(int p[]);
which would tend to inform others who
might use this function that the function is designed to manipulate the
elements of an array. Of course, in either case, what actually gets passed is
the value of a pointer to the first element of the array, independent of which
notation is used in the function prototype or definition. Note that if the
array notation is used, there is no need to pass the actual dimension of the
array since we are not passing the whole array, only the address to the first
element.
We now turn to the problem of the 2
dimensional array. As stated in the last chapter, C interprets a 2 dimensional
array as an array of one dimensional arrays. That being the case, the first
element of a 2 dimensional array of integers is a one dimensional array of
integers. And a pointer to a two dimensional array of integers must be a pointer
to that data type. One way of accomplishing this is through the use of the
keyword "typedef". typedef assigns a new name to a specified data
type. For example:
typedef unsigned char byte;
causes the name byte to mean type
unsigned char. Hence
byte b[10]; would be an array of unsigned characters.
Note that in the typedef declaration,
the word byte has replaced that which would normally be the name of our
unsigned char. That is, the rule for using typedef is that the new name for the
data type is the name used in the definition of the data type. Thus in:
typedef int Array[10];
Array becomes a data type for an array
of 10 integers. i.e. Array my_arr; declares my_arr as an array of 10 integers
and Array arr2d[5]; makes arr2d an array of 5 arrays of 10 integers each.
Also note that Array *p1d; makes p1d a
pointer to an array of 10 integers. Because *p1d points to the same type as
arr2d, assigning the address of the two dimensional array arr2d to p1d, the
pointer to a one dimensional array of 10 integers is acceptable. i.e. p1d =
&arr2d[0]; or p1d = arr2d; are both correct.
Since the data type we use for our pointer is
an array of 10 integers we would expect that incrementing p1d by 1 would change
its value by 10*sizeof(int), which it does. That is, sizeof(*p1d) is 20. You
can prove this to yourself by writing and running a simple short program.
Now,
while using typedef makes things clearer for the reader and easier on the
programmer, it is not really necessary. What we need is a way of declaring a
pointer like p1d without the need of the typedef keyword. It turns out that
this can be done and that int
(*p1d)[10];
is the proper declaration, i.e. p1d
here is a pointer to an array of 10 integers just as it was under the
declaration using the Array type. Note that this is different from
int *p1d[10];
which would make p1d the name of an
array of 10 pointers to type int.
CHAPTER 9:
Pointers and Dynamic Allocation of Memory
There are times when it is convenient
to allocate memory at run time using malloc(), calloc(), or other allocation
functions. Using this approach permits postponing the decision on the size of
the memory block need to store an array, for example, until run time. Or it
permits using a section of memory for the storage of an array of integers at
one point in time, and then when that memory is no longer needed it can be
freed up for other uses, such as the storage of an array of structures.
When memory is allocated, the allocating
function (such as malloc(), calloc(), etc.) returns a pointer. The type of this
pointer depends on whether you are using an older K&R compiler or the newer
ANSI type compiler. With the older compiler the type of the returned pointer is
char, with the ANSI compiler it is void.
If you are using an older compiler, and you
want to allocate memory for an array of integers you will have to cast the char
pointer returned to an integer pointer. For example, to allocate space for 10
integers we might write:
int *iptr;
iptr = (int *)malloc(10 * sizeof(int));
if (iptr == NULL)
{ .. ERROR ROUTINE GOES HERE .. }
If you are using an ANSI compliant
compiler, malloc() returns a void pointer and since a void pointer can be
assigned to a pointer variable of any object type, the (int *) cast shown above
is not needed. The array dimension can be determined at run time and is not
needed at compile time. That is, the 10 above could be a variable read in from
a data file or keyboard, or calculated based on some need, at run time.
Because of the equivalence between array and
pointer notation, once iptr has been assigned as above, one can use the array
notation. For example, one could write:
int k;
for (k = 0; k < 10; k++)
iptr[k] = 2;
to set the values of all elements to
2.
Even with a reasonably good understanding of
pointers and arrays, one place the newcomer to C is likely to stumble at first
is in the dynamic allocation of multi-dimensional arrays. In general, we would
like to be able to access elements of such arrays using array notation, not
pointer notation, wherever possible. Depending on the application we may or may
not know both dimensions at compile time. This leads to a variety of ways to go
about our task.
As we have seen, when dynamically allocating
a one dimensional array its dimension can be determined at run time. Now, when
using dynamic allocation of higher order arrays, we never need to know the
first dimension at compile time. Whether we need to know the higher dimensions
depends on how we go about writing the code. Here I will discuss various
methods of dynamically allocating room for 2 dimensional arrays of integers.
First we will consider cases where the 2nd
dimension is known at compile time.
METHOD 1:
One way of dealing with the problem is
through the use of the typedef keyword. To allocate a 2 dimensional array of
integers recall that the following two notations result in the same object code
being generated:
multi[row][col] = 1; *(*(multi + row) + col) = 1;
It is also true that the following two
notations generate the same code:
multi[row] *(multi + row)
Since the one on the right must
evaluate to a pointer, the array notation on the left must also evaluate to a
pointer. In fact multi[0] will return a pointer to the first integer in the first
row, multi[1] a pointer to the first integer of the second row, etc. Actually,
multi[n] evaluates to a pointer to that array of integers that make up the n-th
row of our 2 dimensional array. That is, multi can be thought of as an array of
arrays and multi[n] as a pointer to the n-th array of this array of arrays.
Here the word pointer is being used to represent an address value. While such
usage is common in the literature, when reading such statements one must be
careful to distinguish between the constant address of an array and a variable
pointer which is a data object in itself.
Consider now:
--------------- Program 9.1 --------------------------------
/* Program 9.1 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
#define COLS 5
typedef int RowArray[COLS];
RowArray *rptr;
int main(void)
{
int nrows = 10;
int row, col;
rptr = malloc(nrows * COLS * sizeof(int));
for (row = 0; row < nrows; row++)
{
for (col = 0; col < COLS; col++)
{
rptr[row][col] = 17;
}
}
return 0;
}
------------- End of Prog. 9.1 --------------------------------
Here I have assumed an ANSI compiler
so a cast on the void pointer returned by malloc() is not required. If you are
using an older K&R compiler you will have to cast using:
rptr = (RowArray *)malloc(.... etc.
Using this approach, rptr has all the
characteristics of an array name name, (except that rptr is modifiable), and
array notation may be used throughout the rest of the program. That also means
that if you intend to write a function to modify the array contents, you must
use COLS as a part of the formal parameter in that function, just as we did
when discussing the passing of two dimensional arrays to a function.
METHOD 2:
In the METHOD 1 above, rptr turned out
to be a pointer to type "one dimensional array of COLS integers". It
turns out that there is syntax which can be used for this type without the need
of typedef. If we write:
int (*xptr)[COLS];
the variable xptr will have all the
same characteristics as the variable rptr in METHOD 1 above, and we need not
use the typedef keyword. Here xptr is a pointer to an array of integers and the
size of that array is given by the #defined COLS. The parenthesis placement
makes the pointer notation predominate, even though the array notation has
higher precedence. i.e. had we written
int *xptr[COLS];
we would have defined xptr as an array
of pointers holding the number of pointers equal to that #defined by COLS. That
is not the same thing at all. However, arrays of pointers have their use in the
dynamic allocation of two dimensional arrays, as will be seen in the next 2
methods.
METHOD 3:
Consider the case where we do not know
the number of elements in each row at compile time, i.e. both the number of
rows and number of columns must be determined at run time. One way of doing
this would be to create an array of pointers to type int and then allocate
space for each row and point these pointers at each row. Consider:
-------------- Program 9.2 ------------------------------------
/* Program 9.2 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 10; /* or read in at run time */
int row;
int **rowptr;
rowptr = malloc(nrows * sizeof(int *));
if (rowptr == NULL)
{
puts("\nFailure to allocate room for row pointers.\n");
exit(0);
}
printf("\n\n\nIndex Pointer(hex) Pointer(dec) Diff.(dec)");
for (row = 0; row < nrows; row++)
{
rowptr[row] = malloc(ncols * sizeof(int));
if (rowptr[row] == NULL)
{
printf("\nFailure to allocate for row[%d]\n",row);
exit(0);
}
printf("\n%d %p %d", row, rowptr[row], rowptr[row]);
if (row > 0)
printf(" %d",(int)(rowptr[row] - rowptr[row-1]));
}
return 0;
}
--------------- End 9.2 -----------------------------------
In the above code rowptr is a pointer
to pointer to type int. In this case it points to the first element of an array
of pointers to type int. Consider the number of calls to malloc():
To get the array of pointers 1 call
To get space for the rows 5 calls
-----
Total 6 calls
If you choose to use this approach
note that while you can use the array notation to access individual elements of
the array, e.g. rowptr[row][col] = 17;, it does not mean that the data in the
"two dimensional array" is contiguous in memory.
You can, however, use the array notation just
as if it were a continuous block of memory. For example, you can write:
rowptr[row][col] = 176;
just as if rowptr were the name of a
two dimensional array created at compile time. Of course row and col must be
within the bounds of the array you have created, just as with an array created
at compile time.
If you want to have a contiguous block of
memory dedicated to the storage of the elements in the array you can do it as
follows:
METHOD 4:
In this method we allocate a block of
memory to hold the whole array first. We then create an array of pointers to
point to each row. Thus even though the array of pointers is being used, the
actual array in memory is contiguous. The code looks like this:
----------------- Program 9.3 -----------------------------------
/* Program 9.3 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int **rptr;
int *aptr;
int *testptr;
int k;
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 8; /* or read in at run time */
int row, col;
/* we now allocate the memory for the array */
aptr = malloc(nrows * ncols * sizeof(int));
if (aptr == NULL)
{
puts("\nFailure to allocate room for the array");
exit(0);
}
/* next we allocate room for the pointers to the rows */
rptr = malloc(nrows * sizeof(int *));
if (rptr == NULL)
{
puts("\nFailure to allocate room for pointers");
exit(0);
}
/* and now we 'point' the pointers */
for (k = 0; k < nrows; k++)
{
rptr[k] = aptr + (k * ncols);
}
/* Now we illustrate how the row pointers are incremented */
printf("\n\nIllustrating how row pointers are incremented");
printf("\n\nIndex Pointer(hex) Diff.(dec)");
for (row = 0; row < nrows; row++)
{
printf("\n%d %p", row, rptr[row]);
if (row > 0)
printf(" %d",(rptr[row] - rptr[row-1]));
}
printf("\n\nAnd now we print out the array\n");
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
rptr[row][col] = row + col;
printf("%d ", rptr[row][col]);
}
putchar('\n');
}
puts("\n");
/* and here we illustrate that we are, in fact, dealing with
a 2 dimensional array in a contiguous block of memory. */
printf("And now we demonstrate that they are contiguous in memory\n");
testptr = aptr;
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
printf("%d ", *(testptr++));
}
putchar('\n');
}
return 0;
}
------------- End Program 9.3 -----------------
Consider again, the number of calls to
malloc()
To get room for the array itself 1 call
To get room for the array of ptrs 1 call
----
Total 2 calls
Now, each call to malloc() creates
additional space overhead since malloc() is generally implemented by the
operating system forming a linked list which contains data concerning the size
of the block. But, more importantly, with large arrays (several hundred rows)
keeping track of what needs to be freed when the time comes can be more
cumbersome. This, combined with the contiguousness of the data block that
permits initialization to all zeroes using memset() would seem to make the
second alternative the preferred one.
As a final example on multidimensional arrays
we will illustrate the dynamic allocation of a three dimensional array. This
example will illustrate one more thing to watch when doing this kind of
allocation. For reasons cited above we will use the approach outlined in
alternative two. Consider the following code:
------------------- Program 9.4 -------------------------------------
/* Program 9.4 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
int X_DIM=16;
int Y_DIM=5;
int Z_DIM=3;
int main(void)
{
char *space;
char ***Arr3D;
int y, z;
ptrdiff_t diff;
/* first we set aside space for the array itself */
space = malloc(X_DIM * Y_DIM * Z_DIM * sizeof(char));
/* next we allocate space of an array of pointers, each
to eventually point to the first element of a
2 dimensional array of pointers to pointers */
Arr3D = malloc(Z_DIM * sizeof(char **));
/* and for each of these we assign a pointer to a newly
allocated array of pointers to a row */
for (z = 0; z < Z_DIM; z++)
{
Arr3D[z] = malloc(Y_DIM * sizeof(char *));
/* and for each space in this array we put a pointer to
the first element of each row in the array space
originally allocated */
for (y = 0; y < Y_DIM; y++)
{
Arr3D[z][y] = space + (z*(X_DIM * Y_DIM) + y*X_DIM);
}
}
/* And, now we check each address in our 3D array to see if
the indexing of the Arr3d pointer leads through in a
continuous manner */
for (z = 0; z < Z_DIM; z++)
{
printf("Location of array %d is %p\n", z, *Arr3D[z]);
for ( y = 0; y < Y_DIM; y++)
{
printf(" Array %d and Row %d starts at %p", z, y, Arr3D[z][y]);
diff = Arr3D[z][y] - space;
printf(" diff = %d ",diff);
printf(" z = %d y = %d\n", z, y);
}
}
return 0;
}
------------------- End of Prog. 9.4 ----------------------------
If you have followed this tutorial up
to this point you should have no problem deciphering the above on the basis of
the comments alone. There are a couple of points that should be made however.
Let's start with the line which reads:
Arr3D[z][y] = space + (z*(X_DIM * Y_DIM) + y*X_DIM);
Note that here space is a character
pointer, which is the same type as Arr3D[z][y]. It is important that when
adding an integer, such as that obtained by evaluation of the expression
(z*(X_DIM * Y_DIM) + y*X_DIM), to a pointer, the result is a new pointer value.
And when assigning pointer values to pointer variables the data types of the value
and variable must match.
CHAPTER 10:
Pointers to Functions
Up to this point we have been
discussing pointers to data objects. C also permits the declaration of pointers
to functions. Pointers to functions have a variety of uses and some of them
will be discussed here.
Consider the following real problem. You want
to write a function that is capable of sorting virtually any collection of data
that can be stored in an array. This might be an array of strings, or integers,
or floats, or even structures. The sorting algorithm can be the same for all.
For example, it could be a simple bubble sort algorithm, or the more complex
shell or quick sort algorithm. We'll use a simple bubble sort for demonstration
purposes.
Sedgewick [1] has described the bubble sort
using C code by setting up a function which when passed a pointer to the array
would sort it. If we call that function bubble(), a sort program is described
by bubble_1.c, which follows:
/*-------------------- bubble_1.c --------------------*/
/* Program bubble_1.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int a[], int N);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int a[], int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (a[j-1] > a[j])
{
t = a[j-1];
a[j-1] = a[j];
a[j] = t;
}
}
}
}
/*---------------------- end bubble_1.c -----------------------*/
The bubble sort is one of the simpler
sorts. The algorithm scans the array from the second to the last element
comparing each element with the one which precedes it. If the one that precedes
it is larger than the current element, the two are swapped so the larger one is
closer to the end of the array. On the first pass, this results in the largest
element ending up at the end of the array. The array is now limited to all
elements except the last and the process repeated. This puts the next largest
element at a point preceding the largest element. The process is repeated for a
number of times equal to the number of elements minus 1. The end result is a
sorted array.
Here our function is designed to sort an
array of integers. Thus in line 1 we are comparing integers and in lines 2
through 4 we are using temporary integer storage to store integers. What we
want to do now is see if we can convert this code so we can use any data type,
i.e. not be restricted to integers.
At the same time we don't want to have to
analyze our algorithm and the code associated with it each time we use it. We
start by removing the comparison from within the function bubble() so as to
make it relatively easy to modify the comparison function without having to
re-write portions related to the actual algorithm. This results in bubble_2.c:
/*---------------------- bubble_2.c -------------------------*/
/* Program bubble_2.c from PTRTUT10.HTM 6/13/97 */
/* Separating the comparison function */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int a[], int N);
int compare(int m, int n);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int a[], int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare(a[j-1], a[j]))
{
t = a[j-1];
a[j-1] = a[j];
a[j] = t;
}
}
}
}
int compare(int m, int n)
{
return (m > n);
}
/*--------------------- end of bubble_2.c -----------------------*/
If our goal is to make our sort
routine data type independent, one way of doing this is to use pointers to type
void to point to the data instead of using the integer data type. As a start in
that direction let's modify a few things in the above so that pointers can be
used. To begin with, we'll stick with pointers to type integer.
/*----------------------- bubble_3.c -------------------------*/
/* Program bubble_3.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int *p, int N);
int compare(int *m, int *n);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int *p, int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare(&p[j-1], &p[j]))
{
t = p[j-1];
p[j-1] = p[j];
p[j] = t;
}
}
}
}
int compare(int *m, int *n)
{
return (*m > *n);
}
/*------------------ end of bubble3.c -------------------------*/
Note the changes. We are now passing a
pointer to an integer (or array of integers) to bubble(). And from within
bubble we are passing pointers to the elements of the array that we want to
compare to our comparison function. And, of course we are dereferencing these
pointer in our compare() function in order to make the actual comparison. Our next
step will be to convert the pointers in bubble() to pointers to type void so
that that function will become more type insensitive. This is shown in
bubble_4.
/*------------------ bubble_4.c ----------------------------*/
/* Program bubble_4.c from PTRTUT10,HTM 6/13/97 */
#include <stdio.h>
int arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(int *p, int N);
int compare(void *m, void *n);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr,10);
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
return 0;
}
void bubble(int *p, int N)
{
int i, j, t;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare((void *)&p[j-1], (void *)&p[j]))
{
t = p[j-1];
p[j-1] = p[j];
p[j] = t;
}
}
}
}
int compare(void *m, void *n)
{
int *m1, *n1;
m1 = (int *)m;
n1 = (int *)n;
return (*m1 > *n1);
}
/*------------------ end of bubble_4.c ---------------------*/
Note that, in doing this, in compare()
we had to introduce the casting of the void pointer types passed to the actual
type being sorted. But, as we'll see later that's okay. And since what is being
passed to bubble() is still a pointer to an array of integers, we had to cast
these pointers to void pointers when we passed them as parameters in our call
to compare().
We now address the problem of what we pass to
bubble(). We want to make the first parameter of that function a void pointer
also. But, that means that within bubble() we need to do something about the
variable t, which is currently an integer. Also, where we use t = p[j-1]; the type
of p[j-1] needs to be known in order to know how many bytes to copy to the
variable t (or whatever we replace t with).
Currently, in bubble_4.c, knowledge within
bubble() as to the type of the data being sorted (and hence the size of each
individual element) is obtained from the fact that the first parameter is a
pointer to type integer. If we are going to be able to use bubble() to sort any
type of data, we need to make that pointer a pointer to type void. But, in
doing so we are going to lose information concerning the size of individual
elements within the array. So, in bubble_5.c we will add a separate parameter
to handle this size information.
These changes, from bubble4.c to bubble5.c
are, perhaps, a bit more extensive than those we have made in the past. So,
compare the two modules carefully for differences.
/*---------------------- bubble5.c ---------------------------*/
/* Program bubble_5.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
long arr[10] = { 3,6,1,2,3,8,4,1,7,2};
void bubble(void *p, size_t width, int N);
int compare(void *m, void *n);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%d ", arr[i]);
}
bubble(arr, sizeof(long), 10);
putchar('\n');
for (i = 0; i < 10; i++)
{
printf("%ld ", arr[i]);
}
return 0;
}
void bubble(void *p, size_t width, int N)
{
int i, j;
unsigned char buf[4];
unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
if (compare((void *)(bp + width*(j-1)),
(void *)(bp + j*width))) /* 1 */
{
/* t = p[j-1]; */
memcpy(buf, bp + width*(j-1), width);
/* p[j-1] = p[j]; */
memcpy(bp + width*(j-1), bp + j*width , width);
/* p[j] = t; */
memcpy(bp + j*width, buf, width);
}
}
}
}
int compare(void *m, void *n)
{
long *m1, *n1;
m1 = (long *)m;
n1 = (long *)n;
return (*m1 > *n1);
}
/*--------------------- end of bubble5.c ---------------------*/
Note that I have changed the data type
of the array from int to long to illustrate the changes needed in the compare()
function. Within bubble() I've done away with the variable t (which we would
have had to change from type int to type long). I have added a buffer of size 4
unsigned characters, which is the size needed to hold a long (this will change
again in future modifications to this code). The unsigned character pointer *bp
is used to point to the base of the array to be sorted, i.e. to the first
element of that array.
We also had to modify what we passed to
compare(), and how we do the swapping of elements that the comparison indicates
need swapping. Use of memcpy() and pointer notation instead of array notation
work towards this reduction in type sensitivity.
Again, making a careful comparison of
bubble5.c with bubble4.c can result in improved understanding of what is happening
and why.
We move now to bubble6.c where we use the
same function bubble() that we used in bubble5.c to sort strings instead of
long integers. Of course we have to change the comparison function since the
means by which strings are compared is different from that by which long
integers are compared. And,in bubble6.c we have deleted the lines within
bubble() that were commented out in bubble5.c.
/*--------------------- bubble6.c ---------------------*/
/* Program bubble_6.c from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <string.h>
#define MAX_BUF 256
char arr2[5][20] = { "Mickey Mouse",
"Donald Duck",
"Minnie Mouse",
"Goofy",
"Ted Jensen" };
void bubble(void *p, int width, int N);
int compare(void *m, void *n);
int main(void)
{
int i;
putchar('\n');
for (i = 0; i < 5; i++)
{
printf("%s\n", arr2[i]);
}
bubble(arr2, 20, 5);
putchar('\n\n');
for (i = 0; i < 5; i++)
{
printf("%s\n", arr2[i]);
}
return 0;
}
void bubble(void *p, int width, int N)
{
int i, j, k;
unsigned char buf[MAX_BUF];
unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
k = compare((void *)(bp + width*(j-1)), (void *)(bp + j*width));
if (k > 0)
{
memcpy(buf, bp + width*(j-1), width);
memcpy(bp + width*(j-1), bp + j*width , width);
memcpy(bp + j*width, buf, width);
}
}
}
}
int compare(void *m, void *n)
{
char *m1 = m;
char *n1 = n;
return (strcmp(m1,n1));
}
/*------------------- end of bubble6.c ---------------------*/
But, the fact that bubble() was unchanged
from that used in bubble5.c indicates that that function is capable of sorting
a wide variety of data types. What is left to do is to pass to bubble() the
name of the comparison function we want to use so that it can be truly
universal. Just as the name of an array is the address of the first element of
the array in the data segment, the name of a function decays into the address
of that function in the code segment. Thus we need to use a pointer to a
function. In this case the comparison function.
Pointers to functions must match the
functions pointed to in the number and types of the parameters and the type of
the return value. In our case, we declare our function pointer as:
int (*fptr)(const void *p1, const void *p2);
Note that were we to write:
int *fptr(const void *p1, const void *p2);
we would have a function prototype for
a function which returned a pointer to type int. That is because in C the
parenthesis () operator have a higher precedence than the pointer * operator.
By putting the parenthesis around the string (*fptr) we indicate that we are
declaring a function pointer.
We now modify our declaration of bubble() by
adding, as its 4th parameter, a function pointer of the proper type. It's
function prototype becomes:
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *));
When we call the bubble(), we insert
the name of the comparison function that we want to use. bubble7.c illustrate
how this approach permits the use of the same bubble() function for sorting
different types of data.
/*------------------- bubble7.c ------------------*/
/* Program bubble_7.c from PTRTUT10.HTM 6/10/97 */
#include <stdio.h>
#include <string.h>
#define MAX_BUF 256
long arr[10] = { 3,6,1,2,3,8,4,1,7,2};
char arr2[5][20] = { "Mickey Mouse",
"Donald Duck",
"Minnie Mouse",
"Goofy",
"Ted Jensen" };
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *));
int compare_string(const void *m, const void *n);
int compare_long(const void *m, const void *n);
int main(void)
{
int i;
puts("\nBefore Sorting:\n");
for (i = 0; i < 10; i++) /* show the long ints */
{
printf("%ld ",arr[i]);
}
puts("\n");
for (i = 0; i < 5; i++) /* show the strings */
{
printf("%s\n", arr2[i]);
}
bubble(arr, 4, 10, compare_long); /* sort the longs */
bubble(arr2, 20, 5, compare_string); /* sort the strings */
puts("\n\nAfter Sorting:\n");
for (i = 0; i < 10; i++) /* show the sorted longs */
{
printf("%d ",arr[i]);
}
puts("\n");
for (i = 0; i < 5; i++) /* show the sorted strings */
{
printf("%s\n", arr2[i]);
}
return 0;
}
void bubble(void *p, int width, int N,
int(*fptr)(const void *, const void *))
{
int i, j, k;
unsigned char buf[MAX_BUF];
unsigned char *bp = p;
for (i = N-1; i >= 0; i--)
{
for (j = 1; j <= i; j++)
{
k = fptr((void *)(bp + width*(j-1)), (void *)(bp + j*width));
if (k > 0)
{
memcpy(buf, bp + width*(j-1), width);
memcpy(bp + width*(j-1), bp + j*width , width);
memcpy(bp + j*width, buf, width);
}
}
}
}
int compare_string(const void *m, const void *n)
{
char *m1 = (char *)m;
char *n1 = (char *)n;
return (strcmp(m1,n1));
}
int compare_long(const void *m, const void *n)
{
long *m1, *n1;
m1 = (long *)m;
n1 = (long *)n;
return (*m1 > *n1);
}
/*----------------- end of bubble7.c -----------------*/
0 Comments