What will be the output of the program?

#include<stdio.h>
#define str(x) #x
#define Xstr(x) str(x)
#define oper multiply

int main()
{
    char *opername = Xstr(oper);
    printf("%s\n", opername);
    return 0;
}

The macro #define Xstr(x) str(x) replaces the symbol 'Xstr(x)' with 'str(x)'.

The macro #define oper multiply replaces the symbol 'oper' with 'multiply'.

Step 1: char *opername = Xstr(oper); The varible *opername is declared as an pointer to a character type.

=> Xstr(oper); becomes,

=> Xstr(multiply);

=> str(multiply)

=> char *opername = multiply

Step 2: printf("%s\n", opername); It prints the value of variable opername.

Hence the output of the program is "multiply"