## operator

[geekocoder]

i compiled the following code and got my output as 100.I am not able to understand the logic behind this...basically the meaning of ## operator...please help me with this

#include <stdio.h>
#define f(g,g2) g##g2
main()
{ int var12=100;
printf("%d",f(var,12));
getchar();
}

[Moschops]

Examine the preprocessor output before compilation. It looks like this:

main()
{ int var12=100;
printf("%d",var12);
getchar();
}

## is not an operator.
f(var,12)
gets turned into
var12

This is a (pretty pointless) preprocessor macro.

[vlad from moscow]

## is a preprocessor directive. It pastes two tokens together. So

f( var, 12 ) is replaced by var12.

After the preprocessing the code will look as

main()
{ int var12=100;
printf("%d",var12);
getchar();
}