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

27 votes

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:

- $100$
- $101$
- $102$
- $110$

41 votes

Best answer

The loop runs $10$ times.

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

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

Correct Answer: $A$

27 votes

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.

3 votes

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