ISRO2015-61

Question

The contents of the flag register after execution of the following program by $8085$ microprocessor will be

$\textbf{Program}$
$\textsf{SUB A}$
$\textsf{MVI B,(01)}_\textsf{H}$
$\textsf{DCR B}$
$\textsf{HLT}$

• A.  $(54)_\textsf{H}$
• B. $(00)_\textsf{H}$
• C. $(01)_\textsf{H}$
• D. $(45)_\textsf{H}$

Comments

please anybody explain this question.I tried allot but unable to understand it.

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m also getting 45h,,,,if v try to use again dcr b...it will result into 95h

Answer 1

My Answer: None of These, ISRO Answer: (a)

Explanation: 8085 microprocessor has a 8 bit flag register to indicate certain conditions arising after arithmetic and logical operations or to control certain operations.

• Carry flag indicates whether there is carry or not after an arithmetic and logical operation.
• Parity flag indicates whether the result contains odd number of 1s or even number of 1s.
• Auxiliary carry (or half carry) flag indicates that is there carry from the 3rd bit to 4th.
• Zero flag indicates whether the result of an arithmetic or logical operation is zero or nonzero.
• Sign flag indicates whether the result is positive or negative.
• Other bits in this register are undefined.

Lets traverse the program:

• SUB A: subtracts the content of accumulator (A) from the content of the accumulator (A); Content of A will become (00)H; Flag Register: 0100 0100 because result is zero and result contains 0 (even) number of 1s.
• MVI B, (01)H : stores 01 in register B; No computation, no change in Flag Register.
• DCR B: decreases content of register B; Content of B will become (00)H; Flag Register: 0100 0100 because result is zero and result contains 0 (even) number of 1s and DCR/INC don't affect the carry flag in 8085. 

So, flag register after execution will be 0100 0100 = (44)H.

Reference: http://labs.kitiyo.com/8085/free-online-assember-simulator-8085.html

 

Comments

Another wrong thing in question - other bits are undefined- so we can't say it is 0.

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Auxilary carry will be 1 after the decrement of (01)H

oooo ooo1 (original B)

1111 1111   (substracting 1) 

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here auxilary carry will be 1 on decrement. so yeah. Isro is correct.

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The answer to this question is (54)H. Here is the proff.

The flag register in 8085 looks like:

S	Z	X	AC	X	P	X	CY
0	0	0	0	0	0	0	0

After performing this operations:

• Instruction 1 will subtract 0 from accumulator register value and initially accumulator register value is 0. So subtracting 0 from 0 will not affect any of the flag except zero flag and parity flag. As final value is zero(even) so Zero flag(Z) will be set and parity flag will be set.
• Instruction 3 is decrementing the value of a register. When initial value of register is 0 and we subtract 1 from it then value of register will become 11111111(considering 8 bit register). It will also set the Auxiliary Cary flag. Sign bit will not be set in decrement operation.

S	Z	X	AC	X	P	X	CY
0	1	0	1	0	1	0	0

 

So the final answer is 0101 0100 and that is (54)H

Answer 2

According to virtual 8085 processor , answer is 45H
please check

Answer 3

ans is a.(54)h

can any one explain iam getting doubt..?

Answer 4

I ran this code in 8085 simulator and step in each step and Got o/p = (45)h as flag value.. Arjun Sir could you explain ...

Comments

I have also doubt in this q? Arjun sir please explain this?

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Are you sure you did the execution correctly? Because then the control should be at HLT instruction. Also, CY at end must be 0 here.

Answer 5

Ans is (54)H OPTION A

Comments

how