# regarding

[jeyshree]

Hai,
I have a doubt.Please do look at the snippet below and answer my question.

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

The output of this program is 100.
What is the function of # here?Please do clear my doubt.Thanks in advance

[whodgson]

the # (pound sign) by itself performs no action and is a null pre-processing directive.
the #include (pound sign ahead of "include")is a pre-processing directive to include the file named in angle brackets(<>)in the translation unit thus replacing the directive with the file.
the ## operator(double pound sign)concatenates arguments. The pre-processor takes the arguments that match the parameter references on either side of the ## operator and converts them to a single token.e.g.

#include<iostream>
#define Amalgamate(a,b,c) a ## b ## c
 
int main()
{  
   unsigned amalgam = Amalgamate(5,12,17);
   std::cout<<amalgam;//prints 51217 to screen
   return 0;
}
 

I've just finished reading about this in Al Stevens book "Teach Yourself C++"!