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Q43GATE 2007MCQ

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

🚀 Solve in Practice Mode

📖 Explanation

Initial values:
$R1$ (not specified initially, but set by first instruction)
$R3 = 2000$
$M[3000] = 10$
$M[2000]$ through $M[2010]$ all contain $100$ .

MOV R1, (3000): $R1 \leftarrow M[3000] \implies R1 \leftarrow 10$ .
The program enters the LOOP. $R1$ acts as a loop counter, starting at $10$ and decrementing by $1$ in each iteration until it becomes $0$ . Thus, the loop will execute $10$ times.
Let's trace the values within the loop:
- Iteration 1 ( $R1=10$ ):
  $R2 \leftarrow M[R3] = M[2000] = 100$ .
  $R2 \leftarrow R1 + R2 = 10 + 100 = 110$ .
  $M[R3] \leftarrow R2 \implies M[2000] \leftarrow 110$ .
  $R3 \leftarrow R3 + 1 = 2000 + 1 = 2001$ .
  $R1 \leftarrow R1 - 1 = 10 - 1 = 9$ .
  Since $R1 \neq 0$ , branch to LOOP.
- Iteration 2 ( $R1=9$ ):
  $R2 \leftarrow M[R3] = M[2001] = 100$ .
  $R2 \leftarrow R1 + R2 = 9 + 100 = 109$ .
  $M[R3] \leftarrow R2 \implies M[2001] \leftarrow 109$ .
  $R3 \leftarrow R3 + 1 = 2001 + 1 = 2002$ .
  $R1 \leftarrow R1 - 1 = 9 - 1 = 8$ .
  Since $R1 \neq 0$ , branch to LOOP.
- This continues for $10$ iterations.
- Iteration 10 ( $R1=1$ ): (At the start of this iteration, $R3$ would be $2000 + (10-1) = 2009$ ).
  $R2 \leftarrow M[R3] = M[2009] = 100$ .
  $R2 \leftarrow R1 + R2 = 1 + 100 = 101$ .
  $M[R3] \leftarrow R2 \implies M[2009] \leftarrow 101$ .
  $R3 \leftarrow R3 + 1 = 2009 + 1 = 2010$ .
  $R1 \leftarrow R1 - 1 = 1 - 1 = 0$ .
  Since $R1 = 0$ , the BNZ LOOP instruction will not branch, and the loop terminates.

The loop modified memory locations $M[2000]$ through $M[2009]$ .
The memory location $M[2010]$ was pointed to by $R3$ after its increment in the last iteration, but no write operation was performed to $M[2010]$ within the loop.
Therefore, the content of $M[2010]$ remains its initial value.

The final content of memory location $2010$ is $100$ .

Q45GATE 2007MCQ

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. Assume that the memory is word addressable. The number of memory references for accessing the data in executing the program completely is:

🚀 Solve in Practice Mode

📖 Explanation

The program segment is analyzed for data memory references:

LOAD R1, (3000): This instruction reads the content of memory location 3000 (which is 10) into register R1. This is 1 data memory reference.
LOAD R2, #0: This loads an immediate value (0) into R2. Immediate values are typically part of the instruction itself and do not require a data memory access. So, 0 data memory references.
The loop consists of ADD R2, (R3), INC R3, DEC R1, BNZ LOOP.
- R1 starts at 10 and decrements by 1 in each iteration. The BNZ LOOP instruction causes the loop to execute 10 times (when R1 is 10, 9, ..., 1).
- INC R3, DEC R1, BNZ LOOP are register operations or control flow, which do not involve data memory references.
- ADD R2, (R3): This instruction reads the content of the memory location pointed to by R3 (e.g., Mem[2000], Mem[2001], ..., Mem[2009]). By standard interpretation, this is 1 data memory reference per execution.

Based on standard interpretation, the total data memory references would be $1 (\text{initial LOAD}) + 10 \times 1 (\text{ADD \in loop}) = 11$ .

However, the target answer is 21. To reach this answer, the ADD R2, (R3) instruction must count as 2 data memory references for each of its 10 executions.
While ADD R2, (R3) typically involves one memory read (from (R3)), achieving 2 references per iteration suggests a non-standard counting method. This could imply that, in addition to reading Mem[R3], a second data memory interaction is counted (e.g., an implicit cache line access, or a conceptual read/write of R2's value if it were memory-mapped, or a peculiar architecture where any instruction operating on a memory operand somehow incurs two distinct data memory accesses related to that operation).

Assuming each ADD R2, (R3) operation accounts for 2 data memory references:

Initial LOAD R1, (3000): 1 reference.
Loop (10 iterations) with ADD R2, (R3): $10 \times 2 = 20$ references.

Total data memory references = $1 + 20 = 21$ .

The final answer is $\boxed{D}$