### Find Output Of Program

Question 6 :

What will be the output of the program?

``````#include
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;
}``````

A). 1, 1, 1, 1
B). 1, 1, 0, 1
C). 1, 0, 0, 1
D). 1, 0, 1, 1

Explanation :

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 7 :

What will be the output of the program?

``````#include
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;
}``````

A). 1, 2, 0, 1
B). -3, 2, 0, 1
C). -2, 3, 0, 1
D). 2, 3, 1, 1

Explanation :

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 8 :

What will be the output of the program?

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

A). 4, 3, 3
B). 4, 3, 2
C). 3, 3, 2
D). 2, 3, 3

Explanation :

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 9 :

What will be the output of the program?

``````#include
int main()
{
int i=3;
i = i++;
printf("%d\n", i);
return 0;
}``````

A). 3
B). 4
C). 5
D). 6

Question 10 :

What will be the output of the program?

``````#include
int main()
{
int a=100, b=200, c;
c = (a == 100 || b > 200);
printf("c=%d\n", c);
return 0;
}``````

A). c=100
B). c=200
C). c=1
D). c=300