Find Output Of Program

Question 11 :

What will be the output of the program?

#include < stdio.h >
#define MIN (x, y) (x 0)
        printf("%d\n", z);
    return 0;
}


A). 3
B). 4
C). 0
D). No output
Answer : Option A

Explanation :

The macro MIN(x, y) (x
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


Question 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;
}


A). Error: in macro substitution
B). Error: invalid reference 'x' in macro
C). print 'multiply'
D). No output
Answer : Option C

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"


Question 13 :

What will be the output of the program?

#include < stdio.h >
#define MESS junk

int main()
{
    printf("MESS\n");
    return 0;
}


A). junk
B). MESS
C). Error
D). Nothing will print
Answer : Option B

Explanation :

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


Question 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;
}


A). 2, 3, 4,
B). 2, 2, 2,
C). 3, 3, 3,
D). 4, 4, 4,
Answer : Option A

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.


Question 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;
}


A). 5
B). 9
C). 10
D). 3+7
Answer : Option C

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".