Understanding the 8086 Assembly Instructions

This solution explains the result of executing a sequence of 8086 assembly instructions, focusing on the final values of the $AL$ register and the Carry Flag ($CF$).

Initial State: MOV Instructions

The process begins with two $MOV$ instructions:

1. $MOV AL, 8EH$: This instruction loads the hexadecimal value $8EH$ into the $AL$ register. In binary representation, $8EH$ is $10001110$.
   • So, after this instruction: $AL$ = $10001110$.
2. $MOV CL, 02H$: This instruction loads the hexadecimal value $02H$ (which is decimal $2$) into the $CL$ register. The $CL$ register typically holds the count for shift or rotate operations.
   • So, after this instruction: $CL$ = $00000010$.

Executing the SAR Instruction

The core of the question lies in the $SAR AL, CL$ instruction. $SAR$ stands for Arithmetic Shift Right.

• Operation: This instruction shifts the bits in the destination operand ($AL$) to the right by the number of positions specified in the source operand ($CL$).
• Key Feature: Arithmetic Shift Right preserves the sign of the number. The value of the Most Significant Bit (MSB) is copied into the Carry Flag ($CF$) and also fills the vacated MSB position(s) on the left.
• Shift Count: The $CL$ register contains $2$, meaning the bits in $AL$ will be shifted right $2$ times.

Step-by-Step Shift Execution:

Initial value of $AL$ is $10001110$. The MSB is $1$. The initial value of $CF$ is irrelevant before the first shift.

Shift 1:

• The MSB of $AL$ ($1$) is shifted into the $CF$ flag. $CF$ becomes $1$.
• All bits in $AL$ shift one position to the right.
• The vacated MSB position is filled with the original MSB ($1$).
• The Least Significant Bit (LSB) ($0$) is discarded.
• $AL$ becomes $11000111$.

Shift 2:

• The MSB of the current $AL$ ($1$) is shifted into the $CF$ flag. $CF$ remains $1$.
• All bits in $AL$ shift one position to the right.
• The vacated MSB position is filled with the original MSB ($1$).
• The LSB ($1$) is discarded.
• $AL$ becomes $11100011$.

Final State

After executing $SAR AL, CL$ with $CL$ = $2$:

• The final value of the $AL$ register is $11100011$.
• The final value of the Carry Flag ($CF$) is $1$.

Therefore, the final state is $AL$ = $11100011$ and $CF$ = $1$. This matches the first option.