Program - 5.a: Add two 16-bit numbers - Source Program 1
Sample problem:
(4000H) = 15H
(4001H) = 1CH
(4002H) = B7H
(4003H) = 5AH
Result = 1C15 + 5AB7H = 76CCH
(4004H) = CCH
(4005H) = 76H
Source Program 1:
LHLD 4000H : Get first I6-bit number in HL
XCHG : Save first I6-bit number in DE
LHLD 4002H : 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 4004H : Store I6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution
Source program 2:
LHLD 4000H : Get first I6-bit number
XCHG : Save first I6-bit number in DE
LHLD 4002H : Get second I6-bit number in HL
DAD D : Add DE and HL
SHLD 4004H : Store I6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution
0 Comments