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]

#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.

 

[parag_paul]

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.

So what is the order of implemenation.
for h(f(1,2))

f(1,2)
happens first
and in the second case,
g(X)
happens first and then f(1,2)

 

[WANG CONG]

So what is the order of implemenation.
for h(f(1,2))

f(1,2)
happens first
and in the second case,
g(X)
happens first and then f(1,2)

See Question 11.17 of C FAQ .

http://c-faq.com/ansi/stringize.html

 

[Eric Sosman]

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

Yes.

which treats its
inputs a strings.

No.

See WANG CONG's response.