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C Preprocessor

Q1: 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;
}

A Error: invalid reference 'x' in macro

B Error: in macro substitution

C No output

D print 'multiply'

ANS:A - Error: in macro substitution

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



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