Question
#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;
}
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;
}
Answer: Option C
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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`setminus`n", opername); It prints the value of variable opername.
Hence the output of the program is "multiply"
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