Move Blocks of Bits in 8085 Assembly Language

Problem Statement − Move a block of 16 data stored from 8050H to 805FH to a target location from 8070F to 807FH.

Address	Hex Codes	Label	Mnemonic	T-States	Comment
8000	21 50 80	START:	LXI H, 8050H	10	Setup HL pair as a pointer for source memory.
8003	11 70 80		LXI D, 8070H	10	Set up DE pair as a pointer for destination memory
8006	06 10		MVI B, 10H	7	Set up B to count 16 bytes
8008	7E	LOOP:	MOV A, M	7	Get data byte from source.
8009	12		STAX D	7	Store data byte as destination
800A	23		INX H	6	Point HL to next source location
800B	13		INX D	6	Point DE to next destination
800C	05		DCR B	4	Decrement count
800D	C2 08 80		JNZ LOOP	10	If counter is not 0, go back to transfer next byte.
8010	76		HLT	5	Stop
Total 17 Bytes				Total 69 T-States

Total T-States = T_{Outside_loop} + T_{Inside_loop}

T_{Outside_loop} = (10+10+7+5) = 32

T_{Inside_loop} = ((7+7+6+6+4+10) * 16 - 3) = (40 * 16 - 3) = 637

so Total T-States = 32 + 637 = 669

The Main program is started from location 8000H - 8010H.

Source Data Block from 8050H - 805FH.

Destination block from 8070H - 807FH

Samual Sam

Updated on: 2019-07-30T22:30:23+05:30

6K+ Views

Get certified by completing the course