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Consider the following program segment. Here $\text{R1, R2}$ and $\text{R3}$ are the general purpose registers.
$$\begin{array}{|c|l|l||c|} \hline & \text {Instruction} & \text{Operation }& \text{Instruction Size} \\ & & & \text{(no. of words)} \\\hline & \text{MOV R1,(3000)} & \text{R1} \leftarrow \text{M[3000]} & \text{2} \\\hline \text{LOOP:}& \text{MOV R2,(R3)} & \text{R2} \leftarrow \text{M[R3]} & \text{1} \\\hline & \text{ADD R2,R1} & \text{R2} \leftarrow \text{R1 + R2} & \text{1} \\\hline & \text{MOV (R3),R2} & \text{M[R3]} \leftarrow \text{R2} & \text{1} \\\hline& \text{INC R3} & \text{R3} \leftarrow \text{R3 + 1} & \text{1} \\\hline & \text{DEC R1} & \text{R1} \leftarrow \text{R1 – 1} & \text{1} \\\hline& \text{BNZ LOOP} & \text{Branch on not zero} & \text{2} \\\hline & \text{HALT} & \text{Stop} & \text{1} \\\hline\end{array}$$

Assume that the content of memory location $3000$ is $10$ and the content of the register $\text{R3}$ is $2000$. The content of each of the memory locations from $2000$ to $2010$ is $100$. The program is loaded from the memory location $1000$. All the numbers are in decimal.

Assume that the memory is word addressable. After the execution of this program, the content of memory location $2010$ is:

1. $100$
2. $101$
3. $102$
4. $110$
edited | 2.2k views
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The loop runs $10$ times.

1. When $R1=10 , \text{Memory}[2000] = 110,$
2. When $R1=9 , \text{Memory}[2001] = 109,$
3. When $R1=8 , \text{Memory}[2002] = 108,$
4. When $R1=7 , \text{Memory}[2003] = 107,$
5. When $R1=6 , \text{Memory}[2004] = 106,$
6. When $R1=5 , \text{Memory}[2005] = 105,$
7. When $R1=4 , \text{Memory}[2006] = 104,$
8. When $R1=3 , \text{Memory}[2007] = 103,$
9. When $R1=2 , \text{Memory}[2008] = 102,$
10. When $R1=1 , \text{Memory}[2009] = 101,$

When $R1=0$ the loop breaks., $\text{Memory}[2010]= 100$

Correct Answer: $A$

edited
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what  is meaning of BNZ ?On which register it should apply ? how did we know it?
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This checks result of prev but one instruction, so R1 applies here
The loop is executed 10 times and it modifies the contents from memory location 2000-2009. Memory location 2010 is untouched - contains 100 as before.
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plz explain...how loop executed??
+1 vote

Ist memory reference R1←M[3000] and then in the loop which runs for 10 times, because the content of memory location 3000 is 10 given in question and loop will run 10 times as

R2← M[R3]

M[R3] ←R2

There are two memory reference every iteration

10*2=20

Total=20+1=21

So  (D) is correct option.

http://www.geeksforgeeks.org/gate-gate-cs-2007-question-71/

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Why loop executed 10 times????

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