The GATE Grind

GATE 2016 EC (EC3) – Question 26

Digital Circuits · Computer Organization · 1 mark · Multiple choice

In an 8085, the accumulator is A7 hexadecimal and the carry flag is 0. After executing RLC, the accumulator and carry respectively are

  1. 4E and 0
  2. 4E and 1
  3. 4F and 0
  4. 4F and 1

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (D) 4F and 1

Explanation

A7 hexadecimal is binary 10100111. RLC rotates all eight accumulator bits left, moving the previous most significant bit into both the least significant bit and carry.

The new bits are 01001111, which is 4F hexadecimal. The old most significant bit was 1, so carry becomes 1. Thus **4F and 1, D**. RLC does not rotate the old carry into the accumulator.

More Computer Organization questions

All Computer Organization questions →