[PRACTICAL] IMPLEMENT PROGRAM TO ADD/SUBTRACT 16 BIT NUMBERS

Are you looking for a step-by-step guide to Implement program to add/subtract 16 bit numbers? here are the steps to implement program to add/subtract 16 bit numbers.

SAMPLE PROGRAM:

LXI H,0101
LXI D,3520
DAD D
HLT
//ANSWER IN HL PAIR 3621

Program: 16 bit addition

(C050H) = 15H

(C051H) = 1CH

(C052H) = B7H

(C053H) = 5AH

Result = 1C15 + 5AB7H = 76CCH

(C054H) = CCH

(C055H) = 76H

Program 1:

LHLD C050H 	: Get first I6-bit number in HL
XCHG 		: Save first I6-bit number in DE
LHLD C052H 	: Get second I6-bit number in HL
MOV A, E 		: Get lower byte of the first number
ADD L 		: Add lower byte of the second number
MOV L, A 		: Store result in L register
MOV A, D 		: Get higher byte of the first number
ADC H 		: Add higher byte of the second number with CARRY
MOV H, A 		: Store result in H register
SHLD C054H 	: Store I6-bit result in memory locations C054H and C0C5H.
HLT 			: Terminate program execution

Program: 16 bit subtraction

(C050H) = 45H

(C051H) = 52H

(C052H) = 14H

(C053H) = 12H

Result = 5245H – 1214H = 4031H

(C054H) = 31H

(C055H) = 40H

Program 2:

LHLD C050H 	: Get first I6-bit number in HL
XCHG 		: Save first I6-bit number in DE
LHLD C052H 	: Get second I6-bit number in HL
MOV A, E 		: Get lower byte of the first number
SUB L 		: Sub lower byte of the second number
MOV L, A 		: Store result in L register
MOV A, D 		: Get higher byte of the first number
SBB H 		: Sub higher byte of the second number with BORROW
MOV H, A 		: Store result in H register
SHLD C054H 	: Store I6-bit result in memory locations C054H and C0C5H.
HLT 			: Terminate program execution

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