Question 1

What will be the output of the program?

#include<stdio.h>
int main()
{
    int x=4, y, z;
    y = --x;
    z = x--;
    printf("%d, %d, %d\n", x, y, z);
    return 0;
}

Answer

Answer: D

Step 1: int x=4, y, z; here variable x, y, z are declared as an integer type and variable x is initialized to 4.
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".

Question 2

What will be the output of the program?

#include<stdio.h>
int main()
{
    int i=-3, j=2, k=0, m;
    m = ++i && ++j || ++k;
    printf("%d, %d, %d, %d\n", i, j, k, m);
    return 0;
}

Answer

Answer: C

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).

Hence the output is "-2, 3, 0, 1".

Question 3

What will be the output of the program?

#include<stdio.h>
int main()
{
    int i=4, j=-1, k=0, w, x, y, z;
    w = i || j || k;
    x = i && j && k;
    y = i || j &&k;
    z = i && j || k;
    printf("%d, %d, %d, %d\n", w, x, y, z);
    return 0;
}

Answer

Answer: D

Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.

Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1

Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0

Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1

Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.

Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".

Question 4

What will be the output of the program?

#include<stdio.h>
int main()
{
    static int a[20];
    int i = 0;
    a[i] = i  ;
    printf("%d, %d, %d\n", a[0], a[1], i);
    return 0;
}

Answer

Answer: C

Step 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will be automatically initialized to value '0'(zero).

Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Step 4: Hence the output is "0, 0, 0".

Question 5

What will be the output of the program?

#include<stdio.h>
int main()
{
    int x=12, y=7, z;
    z = x!=4 || y == 2;
    printf("z=%d\n", z);
    return 0;
}

Answer

Answer: B

Step 1: int x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.

Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.

Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".

Questions for: Expressions