Question

What does the following Java function perform (Assume int occupies four bytes of storage) public static int flint a) Il Pre-conditions a 0 and no overflow/underflow occurs int b 0 for (inti0; i< 32: i++) a a>;// This is a logical shift return b

A. Return the int that represents the number of O's in the binary representation of integer a
B. Return the int that represents the number of 1's in the binary representation of integer a
C. Returns the int that has the binary representation of integer
D. Return the int that has the reversed binary representation of integer a

Answer 1

The given Java function performs a logical right shift on the integer 'a', counting the number of 1's in its binary representation.

Step-by-step explanation:

The given Java function performs a logical right shift on the integer 'a' by 32 times. In each iteration, the shift is performed by 1 bit to the right. The variable 'b' is initialized to 0, and in each iteration, it is updated by adding 1 if the rightmost bit of 'a' is 1.

By the end of the loop, 'b' will represent the number of 1's in the binary representation of 'a'. Therefore, the correct option is B. Return the int that represents the number of 1's in the binary representation of integer 'a'.

The given Java function performs a bitwise operation on the integer a. It initializes integer b to zero and then enters a loop that iterates 32 times, which corresponds to the number of bits in an integer (assuming an int occupies four bytes of storage and each byte has 8 bits).

During each iteration, a is right-shifted one position using the logical shift operator, which does not preserve the sign bit. If the rightmost bit of a is 1 before the shift, then b is incremented. This counts the number of 1s in the binary representation of a. After 32 iterations, the function returns b, which represents the number of 1's in the binary representation of integer a.