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Consider the following 8085 microprocessor program.
LDA 4200 H
CMA
ADI 01 H
STA 4300 H
HLT
If the memory content of 4200 H is 77 H, then the memory content of 4300 H will be -----H.
(Answer up to the nearest integer)
Correct answer is '89'. Can you explain this answer?
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Most Upvoted Answer
Consider the following 8085 microprocessor program.LDA 4200 HCMAADI 0...
Perform 2's complement,
So, 4300 ← 89 H
A← 4200 H
A ← 77 H
CMA → 77 H →10001000→ 88 H
So, 4300 ← 89 HIf the memory content of 4200 H is 77 H, then the memory content of 4300 H will be 89 H.
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Community Answer
Consider the following 8085 microprocessor program.LDA 4200 HCMAADI 0...
Memory Content of 4200 H:
The given program starts with the instruction "LDA 4200 H", which loads the accumulator (A) with the content of memory location 4200 H. According to the question, the memory content at address 4200 H is 77 H.
ALU Operation:
Next, the program executes the instruction "ADI 01 H", which adds 01 H to the accumulator content. The ALU performs the addition operation, and the result is stored back in the accumulator.
Accumulator Content:
Since the initial content of the accumulator is 77 H and 01 H is added to it, the final content of the accumulator after this operation will be 78 H.
Memory Content of 4300 H:
The program then executes the instruction "STA 4300 H", which stores the content of the accumulator into memory location 4300 H. Therefore, the memory content at address 4300 H will be the same as the final content of the accumulator, which is 78 H.
Answer:
According to the question, we need to provide the answer up to the nearest integer. The memory content at address 4300 H is 78 H, which is already an integer. However, the correct answer mentioned is '89'. It appears that there might be an error in the question or answer options provided. Based on the given program and the provided memory content, the memory content at address 4300 H should be 78 H, not 89 H.
In conclusion, the memory content at address 4300 H is 78 H, not 89 H, based on the given program and the provided memory content at address 4200 H.
The given program starts with the instruction "LDA 4200 H", which loads the accumulator (A) with the content of memory location 4200 H. According to the question, the memory content at address 4200 H is 77 H.
ALU Operation:
Next, the program executes the instruction "ADI 01 H", which adds 01 H to the accumulator content. The ALU performs the addition operation, and the result is stored back in the accumulator.
Accumulator Content:
Since the initial content of the accumulator is 77 H and 01 H is added to it, the final content of the accumulator after this operation will be 78 H.
Memory Content of 4300 H:
The program then executes the instruction "STA 4300 H", which stores the content of the accumulator into memory location 4300 H. Therefore, the memory content at address 4300 H will be the same as the final content of the accumulator, which is 78 H.
Answer:
According to the question, we need to provide the answer up to the nearest integer. The memory content at address 4300 H is 78 H, which is already an integer. However, the correct answer mentioned is '89'. It appears that there might be an error in the question or answer options provided. Based on the given program and the provided memory content, the memory content at address 4300 H should be 78 H, not 89 H.
In conclusion, the memory content at address 4300 H is 78 H, not 89 H, based on the given program and the provided memory content at address 4200 H.
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Consider the following 8085 microprocessor program.LDA 4200 HCMAADI 01 HSTA 4300 HHLTIf the memory content of 4200 H is 77 H, then the memory content of 4300 H will be -----H.(Answer up to the nearest integer)Correct answer is '89'. Can you explain this answer?
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