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The contents of the accumulator after the execution of the following program will be
MVI A, C5H
ORA A
RAL
ORA A
RAL
- a)45H
- b)C5H
- c)C4H
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
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The contents of the accumulator after the execution of the following p...
RRC instruction rotate the accumulator right and D0 is placed in D7 .
Most Upvoted Answer
The contents of the accumulator after the execution of the following p...
Explanation:
Given program:
MVI A, C5H
ORA A
RAL
Let's break down the program and explain each instruction:
1. MVI A, C5H: This instruction loads the immediate value C5H into the accumulator (register A). The accumulator is an 8-bit register used for arithmetic and logical operations.
2. ORA A: This instruction performs a logical OR operation between the accumulator (A) and itself. The result is stored back in the accumulator. In this case, the OR operation of any value with itself is the value itself. So, the accumulator remains unchanged.
3. RAL: This instruction rotates the contents of the accumulator left through the carry. The leftmost bit is shifted into the carry flag, and the carry flag is shifted into the rightmost bit of the accumulator.
Now, let's simulate the program execution step by step:
1. Initially, the accumulator contains C5H.
2. The ORA A operation does not change the accumulator value, so it remains C5H.
3. The RAL operation rotates the accumulator left through the carry. Since the accumulator contains C5H, the binary representation is 11000101. After the rotation, the leftmost bit (MSB) is shifted into the carry flag, and the carry flag is shifted into the rightmost bit (LSB). So, the new accumulator value is 10001101, which is 8D in hexadecimal.
Therefore, the final contents of the accumulator after the execution of the program will be 8DH, which corresponds to option A.
Given program:
MVI A, C5H
ORA A
RAL
Let's break down the program and explain each instruction:
1. MVI A, C5H: This instruction loads the immediate value C5H into the accumulator (register A). The accumulator is an 8-bit register used for arithmetic and logical operations.
2. ORA A: This instruction performs a logical OR operation between the accumulator (A) and itself. The result is stored back in the accumulator. In this case, the OR operation of any value with itself is the value itself. So, the accumulator remains unchanged.
3. RAL: This instruction rotates the contents of the accumulator left through the carry. The leftmost bit is shifted into the carry flag, and the carry flag is shifted into the rightmost bit of the accumulator.
Now, let's simulate the program execution step by step:
1. Initially, the accumulator contains C5H.
2. The ORA A operation does not change the accumulator value, so it remains C5H.
3. The RAL operation rotates the accumulator left through the carry. Since the accumulator contains C5H, the binary representation is 11000101. After the rotation, the leftmost bit (MSB) is shifted into the carry flag, and the carry flag is shifted into the rightmost bit (LSB). So, the new accumulator value is 10001101, which is 8D in hexadecimal.
Therefore, the final contents of the accumulator after the execution of the program will be 8DH, which corresponds to option A.
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The contents of the accumulator after the execution of the following program will beMVI A, C5HORA ARALa)45Hb)C5Hc)C4Hd)None of the aboveCorrect answer is option 'A'. Can you explain this answer?
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