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The following program is written for an 8085 microprocessor to add two bytes, located at memory addresses 1FFE and 1FFF
LXIH, 1FFE
MOV B, M
INR L
MOV A, M
ADD B
INR L
MOV M, A
XORA
MOV B, M
INR L
MOV A, M
ADD B
INR L
MOV M, A
XORA
On completion of the execution of the program the result of addition is found.
- a)is the register A
- b)At the memory address 1000
- c)At the memory address 1F0O
- d)At the memory address 2000
Correct answer is option 'C'. Can you explain this answer?
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The following program is written for an 8085 microprocessor to add two...
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Explanation:
To understand why the correct answer is option 'C', let's analyze the program step by step.
Step 1: MOV B, M
This instruction moves the content of the memory location pointed by the HL register pair to the B register. In this case, the memory address 1FFFLXIH is stored in the HL register pair. Therefore, the content of the memory location 1FFFLXIH is moved to the B register.
Step 2: MOV A, M
This instruction moves the content of the memory location pointed by the HL register pair to the A register. As mentioned earlier, the HL register pair contains the memory address 1FFFLXIH. Therefore, the content of the memory location 1FFFLXIH is moved to the A register.
Step 3: ADD B
This instruction adds the content of the B register to the content of the accumulator (A register). The sum is stored in the accumulator. So, the content of the B register is added to the content of the A register.
Step 4: MOV L, M
This instruction moves the content of the memory location pointed by the HL register pair to the L register. As the HL register pair still contains the memory address 1FFFLXIH, the content of the memory location 1FFFLXIH is moved to the L register.
Step 5: MOV M, A
This instruction moves the content of the A register to the memory location pointed by the HL register pair. Since the HL register pair contains the memory address 1FFFLXIH, the content of the A register is moved to the memory location 1FFFLXIH.
Step 6: XRA A
This instruction performs an exclusive OR operation between the content of the A register and itself, resulting in zero. It clears the content of the A register.
Result:
After the execution of the program, the content of the memory location 1FFFLXIH is modified, but the content of the accumulator (A register) remains zero. Therefore, the correct answer is option 'C', which states that the result of the addition is found at the memory address 1F0O.
To understand why the correct answer is option 'C', let's analyze the program step by step.
Step 1: MOV B, M
This instruction moves the content of the memory location pointed by the HL register pair to the B register. In this case, the memory address 1FFFLXIH is stored in the HL register pair. Therefore, the content of the memory location 1FFFLXIH is moved to the B register.
Step 2: MOV A, M
This instruction moves the content of the memory location pointed by the HL register pair to the A register. As mentioned earlier, the HL register pair contains the memory address 1FFFLXIH. Therefore, the content of the memory location 1FFFLXIH is moved to the A register.
Step 3: ADD B
This instruction adds the content of the B register to the content of the accumulator (A register). The sum is stored in the accumulator. So, the content of the B register is added to the content of the A register.
Step 4: MOV L, M
This instruction moves the content of the memory location pointed by the HL register pair to the L register. As the HL register pair still contains the memory address 1FFFLXIH, the content of the memory location 1FFFLXIH is moved to the L register.
Step 5: MOV M, A
This instruction moves the content of the A register to the memory location pointed by the HL register pair. Since the HL register pair contains the memory address 1FFFLXIH, the content of the A register is moved to the memory location 1FFFLXIH.
Step 6: XRA A
This instruction performs an exclusive OR operation between the content of the A register and itself, resulting in zero. It clears the content of the A register.
Result:
After the execution of the program, the content of the memory location 1FFFLXIH is modified, but the content of the accumulator (A register) remains zero. Therefore, the correct answer is option 'C', which states that the result of the addition is found at the memory address 1F0O.
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The following program is written for an 8085 microprocessor to add two bytes, located at memory addresses 1FFE and 1FFFLXIH, 1FFEMOV B, MINR LMOV A, MADD BINR LMOV M, AXORAOn completion of the execution of the program the result of addition is found.a)is the register Ab)At the memory address 1000c)At the memory address 1F0Od)At the memory address 2000Correct answer is option 'C'. Can you explain this answer?
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