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Consider the following program fragment in assembly language of certain hypothetical processor. It has 3 registers, R1, R2 and R3. If R1, R2 and R3 initially contain the values n, 0 and 0 respectively, what is the final value of R3 when the control reaches Z?
X: CMP R1, 0
JZ Z
MOV R2 ,R1 (Copy contents of R1 to R2)
SHR R1 (shift right R1 by 1 bit)
SHL R1
CMP R2, R1
JZ Y
INC R3
Y: SHR R1
JMP X
Z: ......
JZ Z
MOV R2 ,R1 (Copy contents of R1 to R2)
SHR R1 (shift right R1 by 1 bit)
SHL R1
CMP R2, R1
JZ Y
INC R3
Y: SHR R1
JMP X
Z: ......
Correct answer is '0'. Can you explain this answer?
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Most Upvoted Answer
Consider the following program fragment in assembly language of certai...
Let us consider n = 510 = (00000101)2
MOV R2, R1
SHR R1 (00000010)
SHL R1 (00000100)
SHR R1 (00000010)
MOV R2, R1
SHR R1 (00000010)
SHL R1 (00000100)
SHR R1 (00000010)
It counts the number of ones in the binary representation of n.
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Consider the following program fragment in assembly language of certai...
Program Explanation:
The given program fragment is written in assembly language for a hypothetical processor. It uses three registers: R1, R2, and R3. Initially, R1 contains the value 'n', R2 contains the value 0, and R3 contains the value 0.
The program performs the following steps repeatedly until the control reaches the label 'Z':
1. Compare the value in R1 with 0 using the CMP instruction.
2. If the result of the comparison is zero (R1 equals 0), jump to the label 'Z'.
3. If the result of the comparison is not zero, proceed to the next instruction.
4. Copy the contents of R1 to R2 using the MOV instruction.
5. Shift the value in R1 right by 1 bit using the SHR instruction.
6. Shift the value in R1 left by 1 bit using the SHL instruction.
7. Compare the value in R2 with the value in R1 using the CMP instruction.
8. If the result of the comparison is zero (R2 equals R1), jump to the label 'Y'.
9. If the result of the comparison is not zero, proceed to the next instruction.
10. Increment the value in R3 by 1 using the INC instruction.
11. Jump to the label 'Y'.
12. (Label 'Y') Shift the value in R1 right by 1 bit using the SHR instruction.
13. Jump to the label 'X'.
Explanation:
Let's analyze the program step by step to understand its behavior:
X:
1. Compare the value in R1 with 0.
2. If R1 equals 0, jump to label 'Z'.
3. If R1 is not equal to 0:
- Copy the contents of R1 to R2.
- Shift the value in R1 right by 1 bit.
- Shift the value in R1 left by 1 bit.
- Compare the values in R2 and R1.
- If R2 equals R1, jump to label 'Y'.
- If R2 is not equal to R1, proceed to the next instruction.
- Increment the value in R3 by 1.
- Jump to label 'Y'.
Y:
1. Shift the value in R1 right by 1 bit.
2. Jump to label 'X'.
This program fragment essentially performs a bitwise comparison between the original value in R1 (n) and its shifted version. It counts the number of times the shifted value matches the original value and increments R3 accordingly. The program continues this process until the value in R1 becomes 0.
Final Value of R3:
Since the initial value in R1 is 'n' and it is being shifted right by 1 bit in each iteration, eventually, R1 will become 0. This will cause the program to jump to the label 'Z', and the loop will terminate. As R3 is incremented each time the shifted value matches the original value, and there are no matches when R1 becomes 0, the final value of R3 will be 0.
The given program fragment is written in assembly language for a hypothetical processor. It uses three registers: R1, R2, and R3. Initially, R1 contains the value 'n', R2 contains the value 0, and R3 contains the value 0.
The program performs the following steps repeatedly until the control reaches the label 'Z':
1. Compare the value in R1 with 0 using the CMP instruction.
2. If the result of the comparison is zero (R1 equals 0), jump to the label 'Z'.
3. If the result of the comparison is not zero, proceed to the next instruction.
4. Copy the contents of R1 to R2 using the MOV instruction.
5. Shift the value in R1 right by 1 bit using the SHR instruction.
6. Shift the value in R1 left by 1 bit using the SHL instruction.
7. Compare the value in R2 with the value in R1 using the CMP instruction.
8. If the result of the comparison is zero (R2 equals R1), jump to the label 'Y'.
9. If the result of the comparison is not zero, proceed to the next instruction.
10. Increment the value in R3 by 1 using the INC instruction.
11. Jump to the label 'Y'.
12. (Label 'Y') Shift the value in R1 right by 1 bit using the SHR instruction.
13. Jump to the label 'X'.
Explanation:
Let's analyze the program step by step to understand its behavior:
X:
1. Compare the value in R1 with 0.
2. If R1 equals 0, jump to label 'Z'.
3. If R1 is not equal to 0:
- Copy the contents of R1 to R2.
- Shift the value in R1 right by 1 bit.
- Shift the value in R1 left by 1 bit.
- Compare the values in R2 and R1.
- If R2 equals R1, jump to label 'Y'.
- If R2 is not equal to R1, proceed to the next instruction.
- Increment the value in R3 by 1.
- Jump to label 'Y'.
Y:
1. Shift the value in R1 right by 1 bit.
2. Jump to label 'X'.
This program fragment essentially performs a bitwise comparison between the original value in R1 (n) and its shifted version. It counts the number of times the shifted value matches the original value and increments R3 accordingly. The program continues this process until the value in R1 becomes 0.
Final Value of R3:
Since the initial value in R1 is 'n' and it is being shifted right by 1 bit in each iteration, eventually, R1 will become 0. This will cause the program to jump to the label 'Z', and the loop will terminate. As R3 is incremented each time the shifted value matches the original value, and there are no matches when R1 becomes 0, the final value of R3 will be 0.
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Consider the following program fragment in assembly language of certain hypothetical processor. It has 3 registers, R1, R2 and R3. If R1, R2 and R3 initially contain the values n, 0 and 0 respectively, what is the final value of R3 when the control reaches Z?X: CMP R1, 0JZ ZMOV R2 ,R1 (Copy contents of R1 to R2)SHR R1 (shift right R1 by 1 bit)SHL R1CMP R2, R1JZ YINC R3Y: SHR R1JMP XZ: ......Correct answer is '0'. Can you explain this answer?
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