Similarities and differences between arrays and pointers through a practical example

Question

Given the following code:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int a[1];
    int * b = malloc(sizeof(int));

    /* 1 */
    scanf("%d", &a);
    printf("%d\n", a[0]);

    /* 2 */ 
    scanf("%d", &b);
    printf("%d\n", b[0]); 

    return 0;
}

the following warnings are obtained when it is compiled (i686-apple-darwin9-gcc-4.0.1):

array.c: In function 'main':
array.c:9: warning: format '%d' expects type 'int *', but argument 2 has type 'int (*)[0u]'
array.c:14: warning: format '%d' expects type 'int *', but argument 2 has type 'int **'

but, why does an execution-error occur in the second printf, meanwhile it works for the first printf?

Even more, why it is obtained the same output if the first scanf is substituted by scanf("%d", a);?

Thank very much in advance

Answer 1

In most contexts, an expression of array type will be implicitly converted from an "N-element array of T" to "pointer to T" and its value will be set to point to the first element of the array. The exceptions to this rule are when the array is an operand of the & or sizeof operators, or if the array is a string literal being used to initialize another array in a declaration.

So how does all that relate to your code?

In the line

scanf("%d", &a);

You are applying the & operator to the array. This suppresses the implicit conversion from "array of T" to "pointer to T" and returns a value of type "pointer to array of T", or T (*)[N] (hence your first warning). Now it turns out that the value of a pointer to an array and the value of a pointer to the first element of the array are the same, they just have different types. So assuming that a is at address 0x0001000:

expression      type          value         note
----------      ----          -----         ----
         a      int *         0x0001000     implicitly converted to pointer
        &a      int (*)[N]    0x0001000     
     &a[0]      int *         0x0001000

That's why your first call to scanf() "works"; you're passing the right pointer value, but the compiler is complaining because the type of the expression doesn't match what the function expects. Had you written

scanf("%d", a);

you would not have received any warnings, since the type of a will be taken to be int *, which is what scanf() expects. Note that this is identical to calling

scanf("%d", &a[0]);

As for b...

You explicitly declare b as a pointer to int and assign a block of memory to it. When you apply the & operator to it, what you get back is the address of the variable b with type int ** (hence the second warning), not the address that b points to.

expression      type          value         note
----------      ----          -----         ----
         b      int *         0x0040000     value contained in b
        &b      int **        0x0001000     address of b

For that case, you just pass the undecorated b:

scanf("%d", b);