Official PYQ · 2003Q48Machine Instructions and Addressing ModesMCQ1 mark

GATE CS 2003 Q48 — Machine Instructions and Addressing Modes

Consider the following assembly language program for a hypothetical processor $A, B,$ and $C$ are $8-$ bit registers. The meanings of various instructions are shown as comments.
$\small \begin{array}{lll} & \text{MOV B, \#0}&& \text{;} & \text{$B \leftarrow 0$} \\& \text{MOV C, \#8} && \text{;}& \text{$C \leftarrow 8$} \\ \text{Z:} & \text{CMP C, \#0} &&\text{;}& \text{compare C with 0} \\ & \text{JZ X} && \text{;}& \text{jump to X if zero flag is set} \\ & \text{SUB C, \#1} && \text{;}& \text{$C \gets C-1$} \\ & \text{RRC A, \#1} && \text{;}& \text{right rotate A through carry by one bit. Thus:} \\ & \text{} && \text{;}& \text{If the initial values of A and the carry flag are $a\_7\ldots a\_0$ and} \\ & \text{} && \text{;}& \text{$c\_0$ respectively, their values after the execution of this} \\ & \text{} && \text{;}& \text{instruction will be $c\_0a\_7\ldots a\_1$ and $a\_0$ respectively.} \\ & \text{JC Y} && \text{;}& \text{jump to Y if carry flag is set} \\ & \text{JMP Z} && \text{;}& \text{jump to Z} \\ \text{Y:} & \text{ADD B, \#1} && \text{;}& \text{$B \gets B+1$} \\ & \text{JMP Z} && \text{;}& \text{jump to Z} \\ \text{X:}& \text{} && \text{;}& \text{} \\ \end{array}$
If the initial value of register A is A0 the value of register B after the program execution will be

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Answer: B

To determine the final value of register $B$ , let us analyze the step-by-step execution of the assembly program:

Initialization:
- MOV B, #0: Register $B$ is initialized to $0$ . It acts as a counter.
- MOV C, #8: Register $C$ is initialized to $8$ . It acts as a loop counter.
Loop Condition:
- Z: CMP C, #0 followed by JZ X: The loop terminates and jumps to label X when $C = 0$ . Since $C$ starts at $8$ and is decremented by $1$ in each iteration (SUB C, #1), the loop will execute exactly $8$ times.
Rotation and Carry Check:
- RRC A, #1: This instruction rotates register $A$ right through the carry flag by $1$ bit.
  Specifically, the least significant bit (LSB) of $A$ is shifted into the Carry Flag.
- JC Y: If the Carry Flag is $1$ (meaning the shifted-out bit of $A$ was $1$ ), the program jumps to label Y where $B$ is incremented by $1$ (ADD B, #1).
- If the Carry Flag is $0$ , the program bypasses the increment and jumps back to the start of the loop at Z.

Since the loop runs exactly $8$ times, every bit of the $8$ -bit register $A$ (initially containing $A_0$ ) is shifted into the Carry Flag exactly once. Register $B$ is incremented if and only if the shifted bit is $1$ .

Therefore, after the program execution, register $B$ will contain the number of $1$ bits in $A_0$ .

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Published 10 Oct 2026

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