Multiply two eight bit numbers with shift and add method

Statement: Multiply the 8-bit unsigned number in memory location 2200H by the 8-bit unsigned number in memory location 2201H. Store the 8 least significant bits of the result in memory location 2300H and the 8 most significant bits in memory location 2301H.

Sample problem:

(2200) = 1100 (0CH)
(2201) = 0101 (05H)
Multiplicand = 1100 (1210)
Multiplier =  0101 (510)
Result = 12 x 5 = (6010)

Source program : LXI H, 2200 : Initialize the memory pointer MOV E, M : Get multiplicand MVI D, 00H : Extend to 16-bits INX H : Increment memory pointer MOV A, M : Get multiplier LXI H, 0000 : Product = 0 MVI B, 08H : Initialize counter with count 8 MULT: DAD H : Product = product x 2 RAL JNC SKIP : Is carry from multiplier 1 ? DAD D : Yes, Product =Product + Multiplicand SKIP:  DCR B : Is counter = zero JNZ MULT : no, repeat SHLD 2300H : Store the result HLT : End of program

Flowchart for program