Question

What is the content of the accumulator after executing the following code? Answer in decimal format. MOV A,#OABH SETBC RRCA CLR C RLCA

Answer 1

So lets execute the program instruction by instruction

1 Mov A # 0AB H

As data is in hexadecimal so Converting to decimal

0AB = 0000 1010 1011

Data in A = 0000 1010 1011

2. SETB C

This instruction is used to set the Carry Flag so

C=1

3.RRC A

Rotate RIght with carry

In this instruction, each bit is shifted to the adjacent right position. Bit D7 becomes D0. Carry flag C is modified according to the bit D0.

 A = D7 D6 D5 D4 D3 D2 D2 D0

For example

//before the instruction A = 10000001; C=0 //after 1st RRC A = 11000000; C=1 

So RRC A will be

// Before the Instruction we have

A= 0000 1010 1011 ; C = 1

After RRC A

A= 1000 0101 0101 C =1

4. CLR C

This instruction will clear the Carry Flag

C=0

5.

RLC A

In this instruction, each bit is shifted to the adjacent left position. Bit D7 becomes D0. Carry flag C is modified according to the bit D7

For example

//before the instruction A = 10101010; C=0 //after 1st RLC A = 01010101; C=1 

So Now RLC A

// before the instruction

A= 1000 0101 0101 C =0

// After RLC A

A= 1000 0010 1010   C = 1

The final data of accumulator after executing the program

A= 1000 0010 1010 = 82A

In Decimal = 2090

Note ---> Don't confuse IN RRC and RLC with example as that is acc to 8 bit data but our data is 12 bit so we do accordingly.That i only showed so that it is easy to grasp it

Thank You

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