Q1:

What will be the output of the program?

#include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int va1=10;
    int va12=20;
    printf("%d\n", FUN(va1, 2));
    return 0;
}

The following program will make you understand about ## (macro concatenation) operator clearly.

#include<stdio.h>
#define FUN(i, j) i##j

int main()
{
    int First  	= 10;
    int Second  = 20;

    char FirstSecond[] = "AptitudeCrack";

    printf("%s\n", FUN(First, Second) );

    return 0;
}

Output:
-------
AptitudeCrack

The preprocessor will replace FUN(First, Second) as FirstSecond. Therefore, the printf("%s\n", FUN(First, Second) ); statement will become as printf("%s\n", FirstSecond ); Hence it prints AptitudeCrack as output. Like the same, the line printf("%d\n", FUN(va1, 2)); given in the above question will become as printf("%d\n", va12 );. Therefore, it prints 20 as output.