whats happening here

[parag_paul@hotmail.com]

#include <stdio.h>
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)

int main()
{
printf("%s\n",h (f(1,2)));
printf("%s\n",g (f(1,2)));
return 0;
}

[bdsatish]

Re: whats happening here

On Mar 25, 12:23 pm, "parag_p...@hot mail.com" <parag_p...@hot mail.com>
wrote:

#include <stdio.h>
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
>
int main()
{
printf("%s\n",h (f(1,2)));
printf("%s\n",g (f(1,2)));
return 0;
}

f(1,2) ==12 because ## is a concatenating operator, which treats its
inputs a strings. So 1 and 2 are concatenated to form 12

h(a) is simply #a where # is stringisizing operator. So h(12)
becomes the string "12"

Since g(a) is simply #a, the input argument f(1,2) is thrown out as if
it were a string.

[Eric Sosman]

Re: whats happening here

bdsatish wrote:

On Mar 25, 12:23 pm, "parag_p...@hot mail.com" <parag_p...@hot mail.com>
wrote:

>#include <stdio.h>
> #define f(a,b) a##b
> #define g(a) #a
> #define h(a) g(a)
>>
> int main()
> {
> printf("%s\n",h (f(1,2)));
> printf("%s\n",g (f(1,2)));
> return 0;
> }

>
f(1,2) ==12 because ## is a concatenating operator,

Yes.

which treats its
inputs a strings.

No.

See WANG CONG's response.

--
Eric Sosman
esosman@ieee-dot-org.invalid