Topic: Microprocessors
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is
3000 MVI A, 45H
3002 MOV B, A
3003 STC
3004 CMC
3005 RAR
3006 XRA B
- A. 00H
- B. 45H
- C. 67H
- D. E7H
Answer: C. 67H
Explanation: MVI A, 45H loads A with 45H (0100 0101). MOV B, A copies it to B, so B is 45H. STC sets the carry flag (CY=1). CMC complements the carry flag (CY=0). RAR rotates the accumulator right through carry: A becomes 0010 0010 (22H) and CY becomes 1. XRA B performs XOR operation of B with accumulator: 22H XOR 45H = 0010 0010 XOR 0100 0101 = 0110 0111 = 67H.