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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
 (A) 8CH  (B) 64H   (C) 23H   (D) 15H 

Asked On2017-07-06 16:05:29 by:prajwalamv

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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 on:2017-07-09 Answerd By:milan-ransingh

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