Consider the following program segment. Here R1, R2 and R3 are the general…
2007
Consider the following program segment. Here R1, R2 and R3 are the general purpose registers.
Assume that the content of memory location 3000 is 10 and the content of the register 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.
Consider the data given in above question. Assume that the memory is word addressable. After the execution of this program, the content of memory location 2010 is:
Answer: A. 100 — Key idea: determine how many times the loop runs and which memory addresses it updates. Step-by-step execution: MOV R1, (3000) sets R1 = M[3000] = 10. The…
- A.
100
- B.
101
- C.
102
- D.
110
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Show answer & explanation
Correct answer: A
Key idea: determine how many times the loop runs and which memory addresses it updates.
Step-by-step execution:
MOV R1, (3000) sets R1 = M[3000] = 10.
The loop body (for each iteration) does:
R2 ← M[R3] (initially 100), then R2 ← R1 + R2, and M[R3] ← R2.
R3 is incremented by 1 and R1 is decremented by 1.
Because R1 starts at 10, the loop runs exactly 10 times. R3 starts at 2000, so the loop writes to addresses 2000, 2001, …, 2009 (one address per iteration).
Values written on each iteration:
Iteration 1 (R1=10): address 2000 ← 100 + 10 = 110
Iteration 2 (R1=9): address 2001 ← 100 + 9 = 109
Iteration 3 (R1=8): address 2002 ← 100 + 8 = 108
Iteration 4 (R1=7): address 2003 ← 100 + 7 = 107
Iteration 5 (R1=6): address 2004 ← 100 + 6 = 106
Iteration 6 (R1=5): address 2005 ← 100 + 5 = 105
Iteration 7 (R1=4): address 2006 ← 100 + 4 = 104
Iteration 8 (R1=3): address 2007 ← 100 + 3 = 103
Iteration 9 (R1=2): address 2008 ← 100 + 2 = 102
Iteration 10 (R1=1): address 2009 ← 100 + 1 = 101
Since only addresses 2000 through 2009 are updated, memory location 2010 is never modified and therefore retains its original value 100.
Final answer: 100