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Question
The accumulator in 8085 microprocessor contains data 71H and register E contains data 39H. What will be the contents of an accumulator in Hexadecimal after execution of the following instructions independently ?
(i) ADD E
(ii) ORA E
(iii) RRC
Solution
ADD E: ADD E with Accumulator
Before execution : [A] = 71H
instruction : ADD E
71 H = 0 1 1 1 0 0 0 1
ADD 39 H = 0 0 1 1 1 0 0 1
1 0 1 0 1 0 1 0 = AA H
After execution [A] = AA H
[Cy] = 00H
(ii) ORA E : Logically OR with Acc
Before execution : [A] = 71H
Instruction : ORA E
71 H = 0 1 1 1 0 0 0 1
ORA 39 H = 0 0 1 1 1 0 0 1
0 1 1 1 1 0 0 1 = 79 H
After execution : [A] = 79H
RRC : Rotate accumulator right by one bit.
Before execution : [A] = 71H 0 1 1 1 0 0 0 1
Instruction : RRC
After execution [A] = 1 0 1 1 1 0 0 0 = B8 H
∴[A] = B8H
[Cy] = 01H
