GATE CS 2003 Q49 — Machine Instructions and Addressing Modes

Let the initial value of register $A$ be $a_7 a_6 a_5 a_4 a_3 a_2 a_1 a_0$ and the initial state of the carry flag be $C_{in}$.

The loop runs exactly $8$ times (as register $C$ decrements from $8$ to $0$). In each iteration, the RRC A, #1 instruction rotates the bits of $A$ right through the carry flag by $1$ bit:

• Initial state: $A = a_7 a_6 a_5 a_4 a_3 a_2 a_1 a_0$, $\text{Carry} = C_{in}$
• After 1st rotation: $A = C_{in} a_7 a_6 a_5 a_4 a_3 a_2 a_1$, $\text{Carry} = a_0$
• After 2nd rotation: $A = a_0 C_{in} a_7 a_6 a_5 a_4 a_3 a_2$, $\text{Carry} = a_1$
• $\dots$
• After 8th rotation (at label X): $A = a_6 a_5 a_4 a_3 a_2 a_1 a_0 C_{in}$, $\text{Carry} = a_7$

To restore register $A$ to its original value $a_7 a_6 a_5 a_4 a_3 a_2 a_1 a_0$, we need to shift the current carry flag ($a_7$) back into the MSB of $A$, and shift all other bits right by one position.

Executing one more RRC A, #1 at location $X$ does exactly this:

Thus, inserting RRC A, #1 at location $X$ restores the initial value of register $A$.