8085 RLC Instruction Execution

This solution details the outcome of executing the $RLC$ (Rotate Left through Carry) instruction on the 8085 microprocessor, starting from the given accumulator and carry flag states.

Initial Processor State

Before the $RLC$ instruction execution:

• Accumulator (ACC) content: $A7$₁₆
• Carry Flag (CY) status: $0$

Convert the initial accumulator value to its binary representation:

$A7$₁₆ = $1010 0111$₂

RLC Instruction Operation Explained

The $RLC$ instruction performs a left rotation of the accumulator's bits. The key actions are:

• The Most Significant Bit (MSB) of the accumulator is moved into the Carry Flag (CY).
• All other bits of the accumulator are shifted one position to the left.
• The original MSB wraps around and is placed into the Least Significant Bit (LSB) position of the accumulator.

Step-by-Step Rotation

Let's trace the operation using the binary value of the accumulator $1010 0111$:

1. Identify MSB and LSB:
   • Initial ACC: $1010 0111$
   • MSB ($b7$) = $1$
   • LSB ($b0$) = $1$
2. Update Carry Flag: The MSB ($1$) is transferred to the CY flag.
   • New CY = $1$
3. Rotate Accumulator Bits: The bits shift left, with the original MSB wrapping to the LSB position.
   • The bit pattern is $b7 b6 b5 b4 b3 b2 b1 b0$.
   • After rotation, the new pattern becomes $b6 b5 b4 b3 b2 b1 b0 b7$.
   • Applying this to $1010 0111$:
   • $b7$=$1$, $b6$=$0$, $b5$=$1$, $b4$=$0$, $b3$=$0$, $b2$=$1$, $b1$=$1$, $b0$=$1$
   • New ACC bits: $0 1 0 0 1 1 1 1$
   The resulting binary value for the accumulator is $0100 1111$₂.

Final Accumulator and Carry Flag State

Convert the new binary accumulator value back to hexadecimal:

$0100 1111$₂ = $4F$₁₆

Therefore, after the $RLC$ instruction:

• The accumulator content is $4F$₁₆.
• The Carry Flag is $1$.