Passing 2D array as pointer, gcc warning incompatible pointer type

Question

I have the following versions of passing 2D array as pointer.

Version 1

#include <stdio.h>  

void disp(int a[][5])
{
    printf("a[0][3] = %d\n", a[0][3]); /* a[0][3] = 4 */
}

int main ()
{
    int a[10] = {1,2,3,4,5,6,7,8,9,10};
    disp(a);
    return 0;
}

Version 2

#include <stdio.h>

typedef void(*callDisplay)(int*);

void disp(int a[][5])
{
    printf("a[0][3] = %d\n", a[0][3]); /* a[0][3] = 4 */
}

int main ()
{
    int a[10] = {1,2,3,4,5,6,7,8,9,10};
    callDisplay fn = (callDisplay) &disp;
    fn(a);
    return 0;
}

Version 1 rises warning incompatible pointer type. expected int (*)[5] but argument is of type int * as expected. However, (Version 2) calling the same function with pointer is compiling without any such warnings.

gcc options: gcc -O0 -g3 -Wall -c -fmessage-length=0

Could somebody pass light on this?

Veltas · Accepted Answer · 2015-08-11 14:17:27Z

If you remove the cast when assigning the function pointer you get:

tmp.c: In function ‘main’:
tmp.c:13:22: warning: initialization from incompatible pointer type [enabled by default]
     callDisplay fn = &disp;

The cast is suppressing this warning even though by casting to a function pointer of a different type you have invoked undefined behavior when you call the function pointer. Basically, you should never need to cast a function pointer as it will hide any warnings like this.

If you fix the function pointer you get the following code:

typedef void(*callDisplay)(int[][5]);

void disp(int a[][5])
{
    printf("a[0][3] = %d\n", a[0][3]); /* a[0][3] = 4 */
}

int main ()
{
    int a[10] = {1,2,3,4,5,6,7,8,9,10};
    callDisplay fn = &disp;
    fn(a);
    return 0;
}

Which when you compile you get the same warning as your first example:

tmp.c: In function ‘main’:
tmp.c:14:5: warning: passing argument 1 of ‘fn’ from incompatible pointer type [enabled by default]
     fn(a);
     ^
tmp.c:14:5: note: expected ‘int (*)[5]’ but argument is of type ‘int *’

Vlad from Moscow · Accepted Answer · 2015-08-11 13:44:35Z

2

This function declaration

typedef void(*callDisplay)(int*);

has compatible argument when is called like

fn(a);

The problem is related to this casting

callDisplay fn = (callDisplay) &disp;

it is wrong.

That is the program has undefined behaviour.

According to the C Standard (6.3.2.3 Pointers)

8 A pointer to a function of one type may be converted to a pointer to a function of another type and back again; the result shall compare equal to the original pointer. If a converted pointer is used to call a function whose type is not compatible with the referenced type, the behavior is undefined.

answered Aug 11, 2015 at 13:44

Vlad from Moscow

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