Shift a multi-byte BCD number to the right in 8051

Microprocessor8085

Here we will see a problem to shift some multi-byte BCD number to the right. The BCD numbers are shifted by two digits (8-bit). 

Let us consider a four-byte BCD number (45 86 02 78) is stored at location 20H, 21H, 22H,23H. The address 10H holds the number of bytes of the whole BCD number. So after executing this code, the contents will be shifted to the right and 20H will hold 00H.

Address
Value


.
.
.
20H
45
21H
86
22H
02
23H
78






.
.
.

Program

        CLRA;Clear the Register A
        MOVR2,10H;TakeByte Count
        INCR2;Increase R2 for loop
        MOVR1,#20H;Takethe address 20H into R1
LOOP:   XCHA,@R1; Get content, which is pointed out by R1 value
        INCR1;IncreaseR1 for next location
        JNZR2, LOOP ;Check R2 is 0 or not to loop back
HALT:   SJMPHALT ;Stop the program

Here the XCH instruction is used. By using this instruction, the value of register A and content of address pointed by R1 is swapped. Afters wapping, the address is moving one byte to the right, then exchange again to put the old value and take the new value. By using this procedure, the BCD bytes are shifted.

The counter value is increased by 1 at the beginning to run the program one more than the byte count. If the value is not incremented, the last byte will not be shifted, it will be discarded.

Output

Address
Value


.
.
.
20H
00
21H
45
22H
86
23H
02
24H
78


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.
.
raja
Published on 31-Dec-2018 14:14:53
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