Find Output Of Program

Question 6 :

What will be the output of the program, if a short int is 2 bytes wide?

#include<stdio.h>
int main()
{
    short int i = 0;
    for(i<=5 && i>=-1; ++i; i>0)
        printf("%u,", i);
    return 0;
}


A). 1 ... 65535
B). Expression syntax error C. No output D. 0, 1, 2, 3, 4, 5
C). No output
D). 0, 1, 2, 3, 4, 5
Answer : Option A

Explanation :

for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.
In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.
Loop condition always get evaluated to true. Also at this point it increases i by one.
An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)


Question 7 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    char ch;
    if(ch = printf(""))
        printf("It matters\n");
    else
        printf("It doesn't matters\n");
    return 0;
}


A). It matters
B). It doesn't matters
C). matters
D). No output
Answer : Option B

Explanation :

printf() returns the number of charecters printed on the console.

Step 1: if(ch = printf("")) here printf() does not print anything, so it returns '0'(zero).
Step 2: if(ch = 0) here variable ch has the value '0'(zero).
Step 3: if(0) Hence the if condition is not satisfied. So it prints the else statements.
Hence the output is "It doesn't matters".

Note: Compiler shows a warning "possibly incorrect assinment".


Question 8 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    unsigned int i = 65536; /* 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 D

Explanation :

Here unsigned int size is 2 bytes. It varies from 0,1,2,3, ... to 65535.
Step 1:unsigned int i = 65536; here variable i becomes '0'(zero). because unsigned int varies from 0 to 65535.
Step 2: while(i != 0) this statement becomes while(0 != 0). Hence the while(FALSE) condition is not satisfied. So, the inside the statements of while loop will not get executed.
Hence there is no output.

Note: Don't forget that the size of int should be 2 bytes. If you run the above program in GCC it may run infinite loop, because in Linux platform the size of the integer is 4 bytes.


Question 9 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    float a = 0.7;
    if(0.7 > a)
        printf("Hi\n");
    else
        printf("Hello\n");
    return 0;
}


A). Hi
B). Hello
C). Hi Hello
D). None of above
Answer : Option A

Explanation :

if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints 'Hi'
Example:

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}

Output:
0.7000000000 0.6999999881


Question 10 :

What will be the output of the program?

#include<stdio.h>
int main()
{
    int a=0, b=1, c=3;
    *((a) ? &b : &a) = a ? b : c;
    printf("%d, %d, %d\n", a, b, c);
    return 0;
}


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

Explanation :

Step 1: int a=0, b=1, c=3; here variable a, b, and c are declared as integer type and initialized to 0, 1, 3 respectively.
Step 2: *((a) ? &b : &a) = a ? b : c; The right side of the expression(a?b:c) becomes (0?1:3). Hence it return the value '3'.
The left side of the expression *((a) ? &b : &a) becomes *((0) ? &b : &a). Hence this contains the address of the variable a *(&a).
Step 3: *((a) ? &b : &a) = a ? b : c; Finally this statement becomes *(&a)=3. Hence the variable a has the value '3'.
Step 4: printf("%d, %d, %d\n", a, b, c); It prints "3, 1, 3".