UGC-NET CS 2017 December 2nd | Problem 22

Introduction

As a programmer, it is important to constantly update our knowledge and skills. One great resource for doing so is by practicing for competitive exams like UGC-NET CS. In this article, we will discuss the problem 22 from the 2017 December 2nd exam and provide a solution for it.

Problem Statement

Consider the following program:

int main()
{
    int i, x = 32;
    for (i = 0; i < 4; i++)
    {
        x = x << i;
        cout << x << endl;
    }
    return 0;
}

What will be the output of the above program?

Solution

The above program is using bitwise left shift operator (<<) to shift the value of x to the left by i bits in each iteration of the loop. We can see that the value of x is initialized to 32 which is 0b100000 in binary.

In the first iteration, x is shifted left by 0 bits, which means it stays the same. Hence the output on the first line will be 32.

In the second iteration, x is shifted left by 1 bit, which will result in 64 (in binary 0b1000000).

In the third iteration, x is shifted left by 2 bits, which will result in 128 (in binary 0b10000000).

In the fourth and final iteration, x is shifted left by 3 bits, which will result in 256 (in binary 0b100000000).

Therefore, the output of the program will be:

32
64
128
256

Conclusion

By solving competitive programming problems like these, programmers can improve their problem-solving skills and gain a deeper understanding of programming concepts. We hope this article provided some valuable insights into the bitwise left shift operator and how it can be used in programming.