8085 Program to Subtract two 8 Bit numbers

In this program, we will see how to subtract two 8-bit numbers using 8085 microprocessor.

Problem Statement

Write 8085 Assembly language program to subtract two 8-bit numbers and store the result at locations 8050H and 8051H.

Discussion

In 8085, the SUB instruction is used 2’s complemented method for subtraction. When the first operand is larger, the result will be positive. It will not enable the carry flag after completing the subtraction. When the result is negative, then the result will be in 2’s complemented form and carry flag will be enabled.

We are using two numbers at location 8000H and 8001H. When the numbers are 78H and 5DH, then the result will be (78 – 5D = 1B) and when the numbers are 23H and CFH, then the result will be (23 – CF = 154) Here 1 indicates the number is negative. The actual result is 54H. It is in 2’s complemented form.

Input

first input

Address	Data
. . .	. . .
8000	78
8001	5D
. . .	. . .

second input

Address	Data
. . .	. . .
8000	23
8001	CF
. . .	. . .

Flow Diagram

Program

Address	HEX Codes	Labels	Mnemonics	Comments
F000	0E,00		MVIC,00H	Clear C register
F002	21,00, 80		LXIH,8000H	Load initial address to get operand
F005	7E		MOVA, M	Load Acc with the memory element
F006	23		INX H	Point to next location
F007	46		MOVB, M	Load B with the second operand
F008	90		SUB B	Subtract B from A
F009	D2,0D, F0		JNC STORE	When CY = 0, go to STORE
F00C	0C		INR C	Increase C by 1
F00D	21,50, 80	STORE	LXIH,8050H	Load the destination address
F010	77		MOVM, A	Store the result
F011	23		INX H	Point to next location
F012	71		MOVM, C	Store the borrow
F013	76		HLT	Terminate the program