Why stringizing operator is only accepted as a macro?

Question

I tried to use the # operator in my code:

#include <stdio.h>

int main(void)
{
    printf("hello %s !!! n", #world);
    return 0;
}

Compiled with gcc, this codes gives the following error:

str.c: In function ‘main’:
str.c:5:27: error: stray ‘#’ in program
  printf("hello %s !!! n", #world);

But, when I define a macro that uses this operator, the code is compiled:

#include <stdio.h>

#define STRINGIFY(x) #x

int main(void)
{
    printf("hello %s \n", STRINGIFY(world));
    return 0;
}

Is this error reported by the pre-processor? If so, why?

Answer 1

Because #parameter is the part of macro expansion. All this is done by a preprocessor before the compiler even sees the code. And this token is applied to macro parameters only. Consider the following macro:

#define STR(x) #x

Str(Hello) is "Hello"

But if we write #word in the code the preprocessor does not see this as a part of a macro and word is not a macro parameter. Thus the preprocessor ignores it. The compiler sees the same #word and knows nothing what to do with it. So it reports an error. Consider the following text worked out by a preprocessor.

#define STR(x) #x
const char * str=STR(Hello);
const char * buggy_str=#Hello;

The result will be:

const char * str="Hello";
const char * buggy_str=#Hello;

The C compiler sees the first string and it is ok for him. But when he sees the second string, he knows nothing about the token # and thus reports an error.