Code giving compilation error in C99 mode

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?

No, it isn't even valid C89. As the compiler very plainly tells you, you need to convert the pointers to `void *`. — Kerrek SB, Jul 25 '13 at 17:21
@KerrekSB; I have given the link. See the first one, it is compiling! — haccks, Jul 25 '13 at 17:22
http://stackoverflow.com/questions/197757/printing-pointers-in-c — CBIII, Jul 25 '13 at 17:23
The code doesn't violate any syntax rule or constraint, so a conforming C compiler is not required to issue a diagnostic, but it does have undefined behavior. The "all warnings being treated as errors" message means that the compiler is being invoked (probably with `-Werror`) in a way that causes it to treat even optional diagnostics as fatal. All of this is the same for C90, C99, and C11. And a minor point: `int main(void)` is better than `int main()`, though your compiler probably won't care. — Keith Thompson, Jul 25 '13 at 18:10
@KeithThompson; I just tried with `-werror` flag on my compiler (GCC 4.7.1) with C99 mode , and still it is giving the correct output (unlike the second link I have posted). — haccks, Jul 25 '13 at 18:46
@haccks: It's `-Werror`, not `-werror`. `-Werror` only affects the treatment of warnings; it doesn't enable warnings in the first place. This particular warning is enabled by `-pedantic` along with any of `-std=c90`, `-std=c99`, or `-std=c11` (at least for gcc 4.7.2). Add the casts to fix the problem. — Keith Thompson, Jul 25 '13 at 19:06
@KeithThompson; Oops,Sorry! That was `-Werror`. And yes, after enabling `-pedantic` flag it is giving the same compilation error. I was not aware of the use of `-pedantic` flag. Thanks for explanation. — haccks, Jul 25 '13 at 19:58

score 5 · Accepted Answer · answered Jul 25 '13 at 17:29

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);

nice answer, but it don't thought an error for `int*` , whereas `char*` != `int*` — Grijesh Chauhan, Jul 25 '13 at 18:13

Code giving compilation error in C99 mode

1 Answers1