Topic : Pointers and Arrays in C
Author : Ted Jensen
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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 4 byte integers so all copying of rvalues from one storage location to the other is done by copying 4 bytes. Had we been using two byte integers, we would be copying 2 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.  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:


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("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).
References for Chapter 1:
"The C Programming Language" 2nd Edition
B. Kernighan and D. Ritchie
Prentice Hall
ISBN 0-13-110362-8

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, 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, 40 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 4 to ptr instead of 1, so the pointer "points to" the next integer, at memory location 104. Similarly, were the ptr declared as a pointer to a short, it would add 2 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. 4 for an integer).

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];  

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