Code giving compilation error in C99 mode [duplicate]

Question

On testing the code

#include <stdio.h>

 int main()
 {
     char a[5][3];

     printf("a = %p\n", a);
     printf("&a[0] = %p\n", &a[0][0]);
     printf("&a = %p\n", &a);
     printf("*a = %p\n", *a);

     return 0;
  }

it get compiled and giving the output with C mode (http://ideone.com/KD9Wz1):

a = 0xbfd8ea51
&a[0] = 0xbfd8ea51
&a = 0xbfd8ea51
*a = 0xbfd8ea51

While compiling it with strict C99 mode (http://ideone.com/iTACGZ), it results in compilation error:

prog.c: In function ‘main’:
prog.c:7:10: error: format ‘%p’ expects argument of type ‘void *’, but argument 2 has     type ‘char (*)[3]’ [-Werror=format]
prog.c:9:10: error: format ‘%p’ expects argument of type ‘void *’, but argument 2 has    type ‘char (*)[5][3]’ [-Werror=format]
cc1: all warnings being treated as errors

Isn't the above code is valid in C99?

Answer 1

You need to insert an explicit cast to void* for any pointer except char*, because the standard guarantees that character pointers and void pointers have identical representation:

6.2.5.27: A pointer to void shall have the same representation and alignment requirements as a pointer to a character type.

Your C89 program has the same problem, but the compiler does not catch it.

Here is the fix:

printf("a = %p\n", (void*)a);
printf("&a[0] = %p\n", &a[0][0]);
printf("&a = %p\n", (void*)&a);
printf("*a = %p\n", *a);