C Programming - C Preprocessor

1. 

What will be the output of the program?

#include<stdio.h>
#define MAN(x, y) ((x)>(y)) ? (x):(y);

int main()
{
    int i=10, j=5, k=0;
    k = MAN(++i, j++);
    printf("%d, %d, %d\n", i, j, k);
    return 0;
}

A. 12, 6, 12
B. 11, 5, 11
C. 11, 5, Garbage
D. 12, 6, Garbage

Answer: Option A

Explanation:

The macro MAN(x, y) ((x)>(y)) ? (x):(y); returns the biggest number of given two numbers.

Step 1: int i=10, j=5, k=0; The variable i, j, k are declared as an integer type and initialized to value 10, 5, 0 respectively.

Step 2: k = MAN(++i, j++); becomes,

=> k = ((++i)>(j++)) ? (++i):(j++);

=> k = ((11)>(5)) ? (12):(6);

=> k = 12

Step 3: printf("%d, %d, %d\n", i, j, k); It prints the variable i, j, k.

In the above macro step 2 the variable i value is increemented by 2 and variable j value is increemented by 1.

Hence the output of the program is 12, 6, 12


2. 

What will be the output of the program?

#include<stdio.h>
#define SQUARE(x) x*x

int main()
{
    float s=10, u=30, t=2, a;
    a = 2*(s-u*t)/SQUARE(t);
    printf("Result = %f", a);
    return 0;
}

A. Result = -100.000000
B. Result = -25.000000
C. Result = 0.000000
D. Result = 100.000000

Answer: Option A

Explanation:

The macro function SQUARE(x) x*x calculate the square of the given number 'x'. (Eg: 102)

Step 1: float s=10, u=30, t=2, a; Here the variable s, u, t, a are declared as an floating point type and the variable s, u, t are initialized to 10, 30, 2.

Step 2: a = 2*(s-u*t)/SQUARE(t); becomes,

=> a = 2 * (10 - 30 * 2) / t * t; Here SQUARE(t) is replaced by macro to t*t .

=> a = 2 * (10 - 30 * 2) / 2 * 2;

=> a = 2 * (10 - 60) / 2 * 2;

=> a = 2 * (-50) / 2 * 2 ;

=> a = 2 * (-25) * 2 ;

=> a = (-50) * 2 ;

=> a = -100;

Step 3: printf("Result=%f", a); It prints the value of variable 'a'.

Hence the output of the program is -100


3. 

What will be the output of the program?

#include<stdio.h>
#define SQR(x)(x*x)

int main()
{
    int a, b=3;
    a = SQR(b+2);
    printf("%d\n", a);
    return 0;
}

A. 25
B. 11
C. Error
D. Garbage value

Answer: Option B

Explanation:

The macro function SQR(x)(x*x) calculate the square of the given number 'x'. (Eg: 102)

Step 1: int a, b=3; Here the variable a, b are declared as an integer type and the variable b is initialized to 3.

Step 2: a = SQR(b+2); becomes,

=> a = b+2 * b+2; Here SQR(x) is replaced by macro to x*x .

=> a = 3+2 * 3+2;

=> a = 3 + 6 + 2;

=> a = 11;

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

Hence the output of the program is 11


4. 

What will be the output of the program?

#include<stdio.h>
#define JOIN(s1, s2) printf("%s=%s %s=%s \n", #s1, s1, #s2, s2);
int main()
{
    char *str1="India";
    char *str2="BIX";
    JOIN(str1, str2);
    return 0;
}

A. str1=IndiaBIX str2=BIX
B. str1=India str2=BIX
C. str1=India str2=IndiaBIX
D. Error: in macro substitution

Answer: Option B

Explanation:

No answer description available for this question. Let us discuss.

5. 

What will be the output of the program?

#include<stdio.h>
#define CUBE(x) (x*x*x)

int main()
{
    int a, b=3;
    a = CUBE(b++);
    printf("%d, %d\n", a, b);
    return 0;
}

A. 9, 4
B. 27, 4
C. 27, 6
D. Error

Answer: Option C

Explanation:

The macro function CUBE(x) (x*x*x) calculates the cubic value of given number(Eg: 103.)

Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.

Step 2: a = CUBE(b++); becomes

=> a = b++ * b++ * b++;

=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.

=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)

Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.

Hence the output of the program is 27, 6.