[https://www.youtube.com/watch?v=Q6lv42BFKe8:embed:cite]
Based on the sources provided, Question 6 is about converting an 8-bit Gray
code into the corresponding binary code. The function GrayBiCon
implements a well-known conversion technique: you take the original
Gray-coded value and repeatedly XOR it with itself shifted right by one
bit, continuing until the shifted value becomes zero. This works because,
in Gray-to-binary conversion, each binary bit is determined by XORing the
Gray bit with the already-determined higher-order binary bit, so
information must propagate from the most significant side toward the least
significant side. In the program’s structure, the input Gray code is stored
in x, and then both y and z are initialized to x. The variable y
acts as the accumulating result, while z is repeatedly transformed inside
the loop to represent progressively shifted versions of the original
pattern. The loop condition is based on z becoming zero, meaning that
once all shifted contributions have been folded into y, no further
propagation is needed. To achieve this behavior, Blank A must update z by
shifting it right so that each iteration aligns the next higher bit
position with the current position. That operation is the right shift
operator, so A is ＞＞, meaning z ← z ＞＞ 1. Blank B must update y by
XORing it with the shifted value z, because the XOR operation is the
mechanism that merges each shifted contribution into the result. Therefore
B must be y ^ z, meaning y ← y ^ z. The answer key indicates the
correct choice is h, which corresponds exactly to A being ＞＞ and B being
y ^ z. You can see why this is correct by tracing the example where the
input Gray code is 00001100. At initialization, y = 00001100 and z =
00001100. In the first loop pass, z is shifted right one bit to
00000110, then y becomes 00001100 ^ 00000110 = 00001010. In the
second pass, z becomes 00000011, then y becomes 00001010 ^ 00000011
= 00001001. In the third pass, z becomes 00000001, then y becomes
00001001 ^ 00000001 = 00001000. In the fourth pass, z becomes
00000000, then y remains 00001000 after XORing with zero. Now z is
zero, so the loop ends and the function returns y = 00001000, which
matches the expected output described in the sources. What makes this
method particularly elegant is that it avoids explicitly computing each
output bit in isolation. Instead, repeated right shifts ensure that the
influence of higher-order Gray bits is successively propagated downward,
and XOR accumulation ensures those influences combine exactly as the
Gray-to-binary rule requires. The right shift is essential because Gray
conversion depends on “next higher bit” relationships, and the XOR is
essential because Gray code is defined by bitwise differences that are
naturally represented by XOR. That is why option h is the correct answer:
it is the only combination that produces the required
propagation-and-accumulation behavior until the shifted mask becomes zero.
