ITPEC FE Morning October 2020 Question 4

AND with 7F — the mask 0111 1111 clears the parity bit and keeps the lower seven bits as they are.

In an AND, a mask bit of 1 passes the data bit through and a mask bit of 0 forces it to 0. The mask we need is 0 in bit 7 and 1 in bits 0 to 6, which is 0111 1111 = 7F in hexadecimal.

Example: 1010 0110 AND 0111 1111 = 0010 0110.

Answer (b)

Why not others:

• •(a) AND with 0F — 0000 1111 also clears bits 4 to 6, so only the lower four data bits survive
• •(c) OR with 0F — OR cannot clear anything: it forces bits 0 to 3 to 1 and leaves the parity bit untouched
• •(d) XOR with FF — inverts all eight bits; the parity bit is flipped, not removed, and the data bits come out inverted

Key rule: To extract bits, AND with a mask that has 1s at the positions to keep; OR sets bits, XOR inverts them.

Basic Theory · AI-written, may contain errors