+91-6376366224     Send message / ask question on our official What's App Number .

Find Output Of Program

Question 1 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    int i=0;
    for(; i<=5; i++);
        printf("%d", i);
    return 0;
}


A). 0, 1, 2, 3, 4, 5
B). 5
C). 1, 2, 3, 4
D). 6
Answer : Option D

Explanation :

Step 1: int i = 0; here variable i is an integer type and initialized to '0'.
Step 2: for(; i<=5; i++); variable i=0 is already assigned in previous step. The semi-colon at the end of this for loop tells, "there is no more statement is inside the loop".

Loop 1: here i=0, the condition in for(; 0<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 2: here i=1, the condition in for(; 1<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 3: here i=2, the condition in for(; 2<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 4: here i=3, the condition in for(; 3<=5; i++) loop satisfies and then i is increemented by '1'(one)
Loop 5: here i=4, the condition in for(; 4<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 6: here i=5, the condition in for(; 5<=5; i++) loop satisfies and then i is incremented by '1'(one)
Loop 7: here i=6, the condition in for(; 6<=5; i++) loop fails and then i is not incremented.

Step 3: printf("%d", i); here the value of i is 6. Hence the output is '6'. 


Question 2 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    char str[]="C-program";
    int a = 5;
    printf(a >10?"Ps\n":"%s\n", str);
    return 0;
}


A). C-program
B). Ps
C). Error
D). None of above
Answer : Option A

Explanation :

Step 1: char str[]="C-program"; here variable str contains "C-program".
Step 2: int a = 5; here variable a contains "5".
Step 3: printf(a >10?"Ps\n":"%s\n", str); this statement can be written as

if(a > 10)
{
    printf("Ps\n");
}
else
{
    printf("%s\n", str);
}

Here we are checking a > 10 means 5 > 10. Hence this condition will be failed. So it prints variable str.
Hence the output is "C-program".


Question 3 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    int a = 500, b = 100, c;
    if(!a >= 400)
        b = 300;
    c = 200;
    printf("b = %d c = %d\n", b, c);
    return 0;
}


A). b = 300 c = 200
B). b = 100 c = garbage
C). b = 300 c = garbage
D). b = 100 c = 200
Answer : Option D

Explanation :

Initially variables a = 500, b = 100 and c is not assigned.
Step 1: if(!a >= 400)
Step 2: if(!500 >= 400)
Step 3: if(0 >= 400)
Step 4: if(FALSE) Hence the if condition is failed.
Step 5: So, variable c is assigned to a value '200'.
Step 6: printf("b = %d c = %d\n", b, c); It prints value of b and c.
Hence the output is "b = 100 c = 200"


Question 4 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    unsigned int i = 65535; /* Assume 2 byte integer*/
    while(i++ != 0)
        printf("%d",++i);
    printf("\n");
    return 0;
}


A). Infinite loop
B). 0 1 2 ... 65535
C). 0 1 2 ... 32767 - 32766 -32765 -1 0
D). No output
Answer : Option A

Explanation :

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.

Step 1:unsigned int i = 65535;
Step 2:
Loop 1: while(i++ != 0) this statement becomes while(65535 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '1'(variable 'i' is already increemented by '1' in while statement and now increemented by '1' in printf statement) Loop 2: while(i++ != 0) this statement becomes while(1 != 0). Hence the while(TRUE) condition is satisfied. Then the printf("%d", ++i); prints '3'(variable 'i' is already increemented by '1' in while statement and now increemented by '1' in printf statement)
....

The while loop will never stops executing, because variable i will never become '0'(zero). Hence it is an 'Infinite loop'.


Question 5 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    int x = 3;
    float y = 3.0;
    if(x == y)
        printf("x and y are equal");
    else
        printf("x and y are not equal");
    return 0;
}


A). x and y are equal
B). x and y are not equal
C). Unpredictable
D). No output
Answer : Option A

Explanation :

Step 1: int x = 3; here variable x is an integer type and initialized to '3'.
Step 2: float y = 3.0; here variable y is an float type and initialized to '3.0'
Step 3: if(x == y) here we are comparing if(3 == 3.0) hence this condition is satisfied.
Hence it prints "x and y are equal".