C operator questions with answers

C programming tricky objective type operators questions and answers with explanation for written test and interview

(1)

What will be output of the following program?

#include<stdio.h>

int main(){

    float a=0.7;d 

    if(a<0.7){

         printf("C");

    }

    else{

         printf("C++");

    }
    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: c

Turbo C ++4.5: c

Linux GCC: c

Visual C++: c

Explanation: 

The binary value of the double constant 0.7 in 64 bits is 0.1011001100110011001100110011001100110011001100110011. However, when assigned to a floating-point variable 'a' with only 32 bits, 'a' will hold the truncated value 0.10110011001100110011001100110011. As a result, 'a' is less than the original double constant 0.7.

This highlights the impact of precision limitations when representing floating-point numbers with fewer bits than their original precision.

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(2)

What will be output of the following program?

        

#include<stdio.h>

int main(){

    int i=5,j;

    j=++i+++i+++i;

    printf("%d %d",i,j);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 8 24

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error

Explanation:

The expression j = ++i+++i+++i; is equivalent to i = ++i + ++i + ++i;. The initial value of 'i' is 5, and due to the three pre-increment operators, the final value of 'i' becomes 8. This final value is then assigned to the variable 'j', not 'i'. Each pre-increment operation increments 'i' before its value is used in the expression.

So, the result is that 'j' will be assigned the value of 'i' after the three increments, and 'i' will be 8.

So, j=8+8+8

j=24 and

i=8

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(3)

What will be output of the following program?

#include<stdio.h>

int main(){

    int i=1;

    i=2+2*i++;

    printf("%d",i);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 5

Turbo C ++4.5: 5

Linux GCC: 5

Visual C++: 5

Explanation:

When the postfix increment operator (i++) is used in an expression, it first assigns the current value of the variable in the expression and then increments the variable by one.

In the expression i = 2 + 2 * 1, the initial value of 'i' is 2. After the expression is evaluated, 'i' becomes 4. Subsequently, the postfix increment is applied, resulting in 'i' being incremented to 5.

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(4)

What will be output of the following program?

#include<stdio.h>

int main(){

    int a=2,b=7,c=10;

    c=a==b;

    printf("%d",c);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 0

Turbo C ++4.5: 0

Linux GCC: 0

Visual C++: 0

Explanation:

== is a relational operator that returns:

• 1 if the comparison a == b is true
• 0 if the comparison a == b is false

Given the values a = 2 and b = 7, the expression a == b is false (2 is not equal to 7). Therefore, b is assigned the value 0.

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(5)

What will be output of the following program?

#include<stdio.h>

void main(){

    int x;

    x=10,20,30;

    printf("%d",x);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 10

Turbo C ++4.5: 10

Linux GCC: 10

Visual C++: 10

Explanation :

Precedence table:

Operator	Precedence	Associative
=	More than ,	Right to left
,	Least	Left to right

Because the assignment operator (=) has higher precedence than the comma operator, the expression x = 10, 20, 30 is evaluated as follows: first, 10 is assigned to 'x', and then the comma operator is evaluated.

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(6)

What will be output of the following program?

#include<stdio.h>

int main(){

    int a=0,b=10;

    if(a=0){

         printf("true");

    }

    else{

         printf("false");

    }

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: false

Turbo C ++4.5: false

Linux GCC: false

Visual C++: false

Explanation:

The '=' operator is an assignment operator, and 'a = 0' assigns zero to the variable 'a'. In C, zero represents false, and any non-zero number represents true. Therefore, 'if(0)' always evaluates to false, and the code inside the else block will execute.

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(7)

What will be output of the following program?

#include<stdio.h>

int main(){

    int a;

    a=015 + 0x71 +5;

    printf("%d",a);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 131

Turbo C ++4.5: 131

Linux GCC: 131

Visual C++: 131

Explanation:

The octal number 015 represents $5\times 8^0+1\times 8^1=5+8=13$ in decimal. Similarly, the hexadecimal number 0x71 translates to $1\times 16^0+7\times 16^1=1+112=113$ in decimal. Consequently, the expression $�=13+113+5$ yields a value of 131.

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(8)

What will be output of the following program?

#include<stdio.h>

int main(){

    printf("%d %d %d",sizeof(3.14),sizeof(3.14f),sizeof(3.14L));

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 8 4 10

Turbo C ++4.5: 8 4 10

Linux GCC: 8 4 12

Visual C++: 8 4 8

Explanation:

The floating-point constant 3.14f has a size of 4 bytes. The double constant 3.14 (default) has a size of 8 bytes. The long double constant 3.14L has a size of 10 bytes. The sizeof() operator, a keyword in C, returns the size of the data type specified inside the parentheses.

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(9)

What will be output of the following program?

#include<stdio.h>

int main(){

    int x=100,y=20,z=5;

    printf("%d %d %d");

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: 5 20 100

Turbo C ++4.5: 5 20 100

Linux GCC: Garbage values

Visual C++: 5 100 20

By default, variables x, y, and z are of auto type, stored in the stack, which is a Last In First Out (LIFO) data structure. The stack first stores 100, then 20, and finally 5. The program counter points to the top of the stack, which is 5. The default value of %d in printf is the data present in the stack. Therefore, the output is in reverse order of declaration, resulting in the output: 5 20 100.

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(10)

What will be output of the following program?

#include<stdio.h>        

int main(){

    int a=2;

    a=a++ + ~++a;

    printf("%d",a);

    return 0;

}

Explanation

Output: 

Turbo C++ 3.0: -1

Turbo C ++4.5: 0

Linux GCC: 0

Visual C++: 0

Explanation:

Same theory as question (2) and (13).

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(11)

What will be output of the following program?

#include<stdio.h>

int main(){

    int a;

    a=sizeof(!5.6);

    printf("%d",a);

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: 2

Turbo C ++4.5: 2

Linux GCC: 4

Visual C++: 4

Explanation:

The negation operator (!) in C returns an integer value of either 0 or 1. The expression !5.6 is evaluated as follows:

• 1. If the operand is non-zero, ! returns 0.
• 2. If the operand is zero, ! returns 1.

In this case, !5.6 results in 0 because 5.6 is non-zero. It's important to note that the result is of integer type, and the size of the integer data type in this context is typically 4 bytes, not 2 bytes.

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(12)

What will be output of the following program?

#include<stdio.h>

int main(){

    float a;

    (int)a= 45;

    printf("%d,a);

    return 0;

}

Explanation

Output:

Turbo C++ 3.0: Compilation error

Turbo C ++4.5: Compilation error

Linux GCC: Compilation error

Visual C++: Compilation error

Explanation:

After performing any operation on operand it always return some constant value.

(int) i.e. type casting operator is not exception for this. (int) a will return one constant value and we cannot assign any constant value to another constant value in c.

(int)a = 45; is equivalent to
3456 = 45 ( Here 3456 in any garbage value of int(a)).

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(13)

What will be output of the following program?

#include<stdio.h>

int main(){

     int i=5;

     int a=++i + ++i + ++i;

     printf("%d",a);

     return 0;

}

Explanation

Output: 

Turbo C++ 3.0: 21

Turbo C ++4.5: 21

Linux GCC: 22

Visual C++: 24

Explanation:

Rule : ++ (in ++i) is pre increment operator so in any arithmetic expression it first increment the value of variable by one in whole equation up to break point then start assigning the value of variable in the equation. There are many break point operators in. For example:

(1) Declaration statement.
(2) && or operator.
(3) Comma (,) operator etc.

In the following expression:

int a=++i + ++i + ++i;

Here break point is due to declaration .It break after each increment i.e. (initial value of i=5) after first increment value 6 assign to variable i then in next increment will occur and so on.

So, a = 6 + 7 + 8;

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