Question

I have the following program. However, I can’t understand why I have to pass the address of the array. When they are both pointing to the same address. Which is the address of the first element of the array of int’s.

I get a warning when I try and do this “assignment from incompatible pointer type”:

ptr = var;

Complete source code:

void print_values(int (*ptr)[5])
{
    size_t i = 0;
    for(i = 0; i < 5; i++) {
        printf("%d: [ %d ]\n", i, (*ptr)[i]);
    }
}

int main(void)
{
    /* declare a pointer to an array integers */
    int (*ptr)[5] = NULL;
    /* array of integers */
    int var[] = {1, 2, 3, 4, 5};
    /* assign the address of where the array is pointing to (first element) */
    ptr = &var;
    /* Both are pointing to the exact same address */
    printf("var  [ %p ]\n",(void*)var);
    printf("&var [ %p ]\n", (void*)&var);

    print_values(ptr);
    return 0;
}

I compile the code with gcc 4.4.4 c89 -Wall -Wextra -O0

Answer 1

It’s purely a type issue.

In most expression contexts the name of an array (such as var) decays to a pointer to the initial element of the array, not a pointer to the array. [Note that this doesn’t imply that var is a pointer – it very much is not a pointer – it just behaves like a pointer to the first element of the array in most expressions.]

This means that in an expression var normally decays to a pointer to an int, not a pointer to an array of int.

As the operand of the address-of operator (&) is one context where this decay rule doesn’t apply (the other one being as operand of the sizeof operator). In this case the type of &var is derived directly from the type of var so the type is pointer to array of 5 int.

Yes, the pointers have the same address value (the address of an arrays first element is the address of the array itself), but they have different types (int* vs int(*)[5]) so aren’t compatible in the assignment.

ISO/IEC 9899:1999 6.3.2.1/4:

Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type “array of type” is converted to an expression of type “pointer to type” that points to the initial element of the array object and is not an lvalue. …