# C Programming - C Preprocessor

Exercise : C Preprocessor - Find Output of Program
11.
What will be the output of the program?
``````#include<stdio.h>
#define MIN(x, y) (x<y)? x : y;
int main()
{
int x=3, y=4, z;
z = MIN(x+y/2, y-1);
if(z > 0)
printf("%d\n", z);
return 0;
}
``````
3
4
0
No output
Explanation:

The macro MIN(x, y) (x<y)? x : y; returns the smallest value from the given two numbers.

Step 1: int x=3, y=4, z; The variable x, y, z are declared as an integer type and the variable x, y are initialized to value 3, 4 respectively.

Step 2: z = MIN(x+y/2, y-1); becomes,

=> z = (x+y/2 < y-1)? x+y/2 : y - 1;

=> z = (3+4/2 < 4-1)? 3+4/2 : 4 - 1;

=> z = (3+2 < 4-1)? 3+2 : 4 - 1;

=> z = (5 < 3)? 5 : 3;

The macro return the number 3 and it is stored in the variable z.

Step 3: if(z > 0) becomes if(3 > 0) here the if condition is satisfied. It executes the if block statements.

Step 4: printf("%d\n", z);. It prints the value of variable z.

Hence the output of the program is 3

12.
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;
}
``````
Error: in macro substitution
Error: invalid reference 'x' in macro
print 'multiply'
No output
Explanation:

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"

13.
What will be the output of the program?
``````#include<stdio.h>
#define MESS junk

int main()
{
printf("MESS\n");
return 0;
}
``````
junk
MESS
Error
Nothing will print
Explanation:

printf("MESS\n"); It prints the text "MESS". There is no macro calling inside the printf statement occured.

14.
What will be the output of the program?
``````#include<stdio.h>
#define PRINT(i) printf("%d,",i)

int main()
{
int x=2, y=3, z=4;
PRINT(x);
PRINT(y);
PRINT(z);
return 0;
}
``````
2, 3, 4,
2, 2, 2,
3, 3, 3,
4, 4, 4,
Explanation:

The macro PRINT(i) print("%d,", i); prints the given variable value in an integer format.

Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.

Step 2: PRINT(x); becomes printf("%d,",x). Hence it prints '2'.

Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'.

Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'.

Hence the output of the program is 2, 3, 4.

15.
What will be the output of the program?
``````#include<stdio.h>
#define MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c)

int main()
{
int x;
x = MAX(3+2, 2+7, 3+7);
printf("%d\n", x);
return 0;
}
``````
5
9
10
3+7
Explanation:

The macro MAX(a, b, c) (a>b ? a>c ? a : c: b>c ? b : c) returns the biggest of given three numbers.

Step 1: int x; The variable x is declared as an integer type.

Step 2: x = MAX(3+2, 2+7, 3+7); becomes,

=> x = (3+2 >2+7 ? 3+2 > 3+7 ? 3+2 : 3+7: 2+7 > 3+7 ? 2+7 : 3+7)

=> x = (5 >9 ? (5 > 10 ? 5 : 10): (9 > 10 ? 9 : 10) )

=> x = (5 >9 ? (10): (10) )

=> x = 10

Step 3: printf("%d\n", x); It prints the value of 'x'.

Hence the output of the program is "10".