GATE Overflow Test Series | Programming | Test 1 | Question: 25

Question

Match the following expressions with the functions they are doing:

$
\begin{array}{|cc|cl|}
\hline
 A. & \text{!(n & n-1)} &1.&  \text{Sets the }{(d+1})^{th}\text{ bit from right of n to 1}\\ \hline
B. &  \text{n | (1<<d)}&2.& \text{Unsets the }{(d+1)}^{th}\text{  bit from right of n to 0}\\ \hline 
C. &  \text{n & (1<<d)}&3.& \text{Checks if n is a power of 2}\\ \hline
D. &  \text{ n ^ (1<<d)}&4.& \text{Complements the }{(d+1)}^{th}\text{  bit from right of n}\\ \hline 
 &  &5.& \text{Something else}\\ \hline   
\end{array}
$

• A. $A-3, B-1, C-2, D-4$
• B. $A-3, B-2, C -1, D-4$
• C. $A-3, B-1, C-5, D-4$
• D. $A-5, B-2, C-1, D- 3$

 

Answer 1

C is the correct answer.

$1 << d$ returns a number with ${(d+1)}^{th}$ bit from right side set to $1.$ So, bitwise OR with this will set the ${(d+1)}^{th}$ bit and bitwise XOR will complement it. But bitwise AND will also affect the other bits and it is not doing the unset operation.

if $n$ is a power of $2$ it'll have just one $“1"$ and all bits to its right will be $0's.$ And $n-1$ will have $1's$ in all those bits corresponding to the $0's$ in $n.$ So, $n\&(n-1)$ returns $0$ for any power of $2.$