What will be the content in register A, after execution of following code in 8051? (Suffix H denotes hexadecimal number)

CLR A

MOV RO, #77H

CPL A

XRL A, RO

MOV A, RO

Explanation:

Step-by-Step Execution:

1. CLR A:

The "CLR A" instruction clears the accumulator register A, setting its content to 00H. After this instruction, the value of register A becomes:

A = 00H

2. MOV R0, #77H:

The "MOV R0, #77H" instruction moves the immediate value 77H into register R0. Now, the value of register R0 becomes:

R0 = 77H

3. CPL A:

The "CPL A" instruction complements the contents of register A bit by bit. Since A = 00H (all bits are 0), complementing it will result in FFH (all bits are 1). After this instruction, the value of register A becomes:

A = FFH

4. XRL A, R0:

The "XRL A, R0" instruction performs a bitwise XOR operation between the contents of register A and register R0. XOR operation rules are as follows:

• 0 XOR 0 = 0
• 0 XOR 1 = 1
• 1 XOR 0 = 1
• 1 XOR 1 = 0

Currently:

A = FFH (binary: 11111111)

R0 = 77H (binary: 01110111)

Performing XOR, we get:

11111111 (A)

01110111 (R0)

Result: 10001000 (binary) = 88H

After this instruction, the value of register A becomes:

A = 88H

5. MOV A, R0:

The "MOV A, R0" instruction moves the contents of register R0 into register A. Now, the value of register A becomes:

A = 77H

The correct answer is: Option 3: 77H