whats happening here

[parag_paul]

#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]

On Mar 25, 12:23 pm, "[e-mail]" <[e-mail]>
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]

bdsatish wrote:

On Mar 25, 12:23 pm, "[e-mail]" <[e-mail]>
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
es*****@ieee-dot-org.invalid