Subtract the 16-bit number in memory locations 4002H and 4003H from the 16-bit number in memory locations 4000H and 4001H. The most significant eight

Sample problem

(4000H) = 19H

(400IH) = 6AH

(4004H) = I5H (4003H) = 5CH

Result = 6A19H - 5C15H = OE04H

(4004H) = 04H

(4005H) = OEH

Source program:

LHLD 4000H : Get first 16-bit number in HL

XCHG : Save first 16-bit number in DE

LHLD 4002H : Get second 16-bit number in HL

MOV A, E : Get lower byte of the first number

SUB L : Subtract lower byte of the second number

MOV L, A : Store the result in L register

MOV A, D : Get higher byte of the first number

SBB H : Subtract higher byte of second number with borrow

MOV H, A : Store l6-bit result in memory locations 4004H and 4005H.

SHLD 4004H : Store l6-bit result in memory locations 4004H and 4005H.

HLT : Terminate program execution.