The content of the accumulator after the execution of the following 8085…
2016
The content of the accumulator after the execution of the following 8085 assembly language program, is
MVI A, 35H
MOV B, A
STC
CMC
RAR
XRA B
Answer: D. 2FH — Solution: Simulate the program step by step. MVI A, 35H -> A = 35H (00110101b). MOV B, A -> B = 35H. STC sets Carry = 1. CMC complements Carry -> Carry = 0.…
- A.
00H
- B.
35H
- C.
EFH
- D.
2FH
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Show answer & explanation
Correct answer: D
Solution: Simulate the program step by step.
MVI A, 35H -> A = 35H (00110101b). MOV B, A -> B = 35H.
STC sets Carry = 1. CMC complements Carry -> Carry = 0.
RAR (rotate A right through Carry): before RAR A = 00110101b and Carry = 0. RAR makes A = (Carry << 7) | (A >> 1) = 00011010b = 1AH, and the old A0 (1) becomes Carry = 1.
XRA B (A = A XOR B): A = 1AH (00011010b) XOR B = 35H (00110101b) -> result = 00101111b = 2FH.
Final accumulator value: 2FH