Instruction Set and Programming with 8085 · Microprocessors · GATE ECE
Marks 1
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From which address will be text instruction be fetched?
1000: LXI SP< 27FF
1003: CALL 1006
1006: POP H
The contents of the stack pointer (SP) and the HL register pair on completion or execution of these instructions are.
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Marks 2
MVI A, 33H
MVI B, 78H
ADD B
CMA
ANI 32H
The Accumulator value immediately after the execution of the fifth instruction is
MVI A,07H
RLC
MOV B,A
RLC
RLC
ADD B
RRC
3000 MVI A, 45H
3002 MOV B, A
3003 STC
3004 CMC
3005 RAR
3006 XRA B
2710 LXI H, 30A0H
DAD H
PCHL
All addresses and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL.
Which of the following statements is correct.Line 1: MVI A, B5H
2: MVI B, 0EH
3: XRI 69H
4: ADD B
5: ANI 9BH
6: CPI 9FH
7: STA 3010H
8: HLT
The contents of the accumulator just after execution of the ADD instruction in line 4 will be
Line 1: MVI A, B5H
2: MVI B, 0EH
3: XRI 69H
4: ADD B
5: ANI 9BH
6: CPI 9FH
7: STA 3010H
8: HLT
After execution of line of the program, the status of the CY and Z flags will be
LXI SP, EFFF H
CALL 3000 H
3000H: LXI H, 3CF4H
PUSH PSW
SPHL
POP PSW
RET
On completion of RET execution, the contents of SP is
Consider an 8085 microprocessor system
The following program starts at locaton 0100H.
LXI SP, 00FF
LXI H, 0107
MVI A, 20H
SUB M
The contents of accumulator wnen the program counter reaches 0109H is
Consider an 8085 microprocessor system
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
LXI H, 9258
MOV A, M
CMA
MOV M, A
Which one of the following is performed by this sequence?
MVI A, 00H
LOOP:.........
...............
HLT END
The sequence of instruction to be complete the program would be
(I.) LDA 3000H
(II.) LXI D, FOF 1H
Would be
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Marks 5
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LOC
2000 START: LXI SP, 1000H
LXI H, 2F37 H
XRA A
MOV A, H
INX H
PUSH H
CZ 20 FF H JMP 3000 H
HLT
20FF ADD H
RZ
POP B
PUSH B
RNZ
HLT