An 8085 assembly language program is given below gate ece 2011

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GATE-ECE-2011--->View question

An 8085 assembly language program is given below. -gate-ece-2011

Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is
MVI A,07H
RLC
MOV B,A
RLC
RLC
ADD B
RRC
 (A) 8CH  (B) 64H   (C) 23H   (D) 15H 



Asked by:prajwalamv

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Answers
MVI A, 35H   // content of accumulator = 35H = 0011 0101

MOV B, A     // content of register B = 35H

STC              // set carry flag = 1.

CMC             // complement carry flag = not( 1) = 0.

RAR              // Each binary bit of the accumulator is rotated right by one position through the Carry flag. And modify carry = D0.

and D7 = previous carry.

                      //  Now content of accumulator = 0001 1010  , carry = 1.

XRA B             // The content of accumulator are exclusive OR with register B. 

 (0011 0101 ) xor ( 0001 1010 ) 

 = 0010 1111

 = 2F H ( Ans - D)

Answerd By:milan-ransingh

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