GATE CSE 2007 | Question: 72

Question

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

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:

• A. $100$
• B. $101$
• C. $102$
• D. $110$

Comments

Next question:
https://gateoverflow.in/43516/gate2007-73

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Memory location 2010 remains untouched as at memory location 2009 , the loop breaks as BNZ LOOP becomes FALSE. So, M[2009] = 101 and M[2010] = 100{remains untouched till end}.

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Similar:

https://gateoverflow.in/357397/Gate-cse-2021-set-1-question-55

Answer 1

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: $A$

Comments

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

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Good question.

Answer 2

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.

Comments

plz explain...how loop executed??

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The line memory location 2010 remains untouched is enough and must have been added to the best answer. Thanks :)

Answer 3

Answer : A

R1 = 10

initial 2000-2010

100

100

100

100

100

100

100

100

100

100

100

2000     2001    2002    2003    2004     2005        2006       2007    2008      2009     2010

1. R1 = 10 , R2 = M[R3]= 100 , R2=R1+R2=110 , R3=2000 , M[2000] = R2=110 , R3 = 2000+1 = 2001

2. R1 = 9 , R2 = M[R3]= 100 , R2=R1+R2=109 , R3=2001 , M[2001] = R2=109 , R3 = 2001 + 1 = 2002 

3. R1 = 8 , R2 = M[R3]= 100 , R2=R1+R2=108 , R3=2002 , M[2002] = R2=108 ,  R3 = 2002 + 1 = 2003 

4. R1 = 7 , R2 = M[R3]= 100 , R2=R1+R2=107 , R3=2003 , M[2003] = R2=107 , R3 = 2003 + 1 = 2004  

5. R1 = 6 , R2 = M[R3]= 100 , R2=R1+R2=106 , R3=2004 , M[2004] = R2=106  , R3 = 2004 + 1 = 2005 

6. R1 = 5 , R2 = M[R3]= 100 , R2=R1+R2=105 , R3=2005 , M[2005] = R2=105   , R3 = 2005 + 1 = 2006 

7. R1 = 4 , R2 = M[R3]= 100 , R2=R1+R2=104 , R3=2006 , M[2006] = R2=104   , R3 = 2006 + 1 = 2007 

8. R1 = 3 , R2 = M[R3]= 100 , R2=R1+R2=103 , R3=2007 , M[2007] = R2=103   , R3 = 2008 + 1 = 2009 

9. R1 = 2 , R2 = M[R3]= 100 , R2=R1+R2=102 , R3=2008 , M[2008] = R2=102    , R3 = 2009 + 1 = 2010 

10. R1 = 1, R2 = M[R3]= 100 , R2=R1+R2=101 , R3=2009 , M[2009] = R2=101 , R3 = 2010 + 1 = 2011

10. R1 = 0 HALT

Final

110

109

108

107

106

105

104

103

102

101

100

 2000   2001  2002 2003    2004   2005      2006      2007      2008      2009      2010

 

Answer 4

Answer is 100.because loop is terminated when address is 2009