GATE CS 2021 Set 1 Q55 — Machine Instructions and Addressing Modes

To find the content of memory location $3010$ after execution, let us first determine the instruction addresses based on their sizes starting from address $1000$:

• 1000: MOV R1, (5000) (4 bytes)
• 1004: MOV R2, (R3) (4 bytes)
• 1008: ADD R2, R1 (2 bytes)
• 1010: MOV (R3), R2 (4 bytes)
• 1014: INC R3 (2 bytes)
• 1016: DEC R1 (2 bytes)
• 1018: BNZ 1004 (2 bytes)
• 1020: HALT (1 byte)

Before the loop starts, we have:

• $R1 = \text{MEMORY}[5000] = 10$
• $R3 = 3000$

The loop at address $1004$ executes as long as $R1 \neq 0$. In each iteration $k$ (from $1$ to $10$):

1. The value at $\text{MEMORY}[R3]$ is read into $R2$.
2. $R2$ is updated to $R1 + R2$.
3. The updated value of $R2$ is written back to $\text{MEMORY}[R3]$.
4. $R3$ is incremented by $1$.
5. $R1$ is decremented by $1$.

Tracing the memory updates and register values at the end of each iteration:

• Iteration 1: Updates $\text{MEMORY}[3000]$. $R3$ becomes $3001$, $R1$ becomes $9$.
• Iteration 2: Updates $\text{MEMORY}[3001]$. $R3$ becomes $3002$, $R1$ becomes $8$.
• $\dots$
• Iteration 10: Updates $\text{MEMORY}[3009]$. $R3$ becomes $3010$, $R1$ becomes $0$.

At the end of the $10^{\text{th}}$ iteration, since $R1 = 0$, the loop condition for BNZ 1004 fails, and the program halts. Thus, the memory location $3010$ is never modified, and its content remains at its initial value of 50.