Got this from a GATE paper of 2005
Im confused. any clue?


The following program starts at location 0100H
LXI SP, 00FF
LXI H, 0701
MVI A, 20H
SUB M

The content od accumulator when the program counter reaches 0109H is
(A) 20H (B)02H (C)00H (D)FFH

If in addition following code exists from 0109H onwards
ORI 40H
ADD M

what will be the result in the accumulator after the last instruction is executed?
(A)40H (B) 20H (C) 60H (D) 42H

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I don't know that it's about, clarify!

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You have to choose the right option from a, b, c, d

GATE is graduate aptitude test for engineers , an exam conducted in India for entrence to postgraduate courses

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If you assume that the program is assembled with assembler as shown then the information is incomplete because
To execute the instruction SUB M at location 0108 you need the content of memory location 0701 pointed by [HL] to perform the operaion SUB M which is A = A-[HL]

Assuming that the data apears in memory as shown in the instructions which means
0100 LXI SP
0101 00
0102 FF
0103 LXI H
0104 07
0105 01
0106 MVI A
0107 20
0108 SUB M

if the memory content is as above then
the answer will be option C

The answer is option C

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