Sample Problem: (6000) H = 8AH
1.8AH ? 64H (Decimal 100) :. Divide by 64H (Decimal 100)
8AH/64H ? Quotient = 1, Remainder = 26H
26H < 64H (Decimal 100) :. Go to step 2 and Digit 2 = 1
2.26H ? OAH (Decimal 10) :. Divide by OAH (Decimal 10)
26H/OAH ? Quotient = 3, Remainder = O8H
OSH < OAH (Decimal 10) :. Go to step 3 and Digit 1 = 3
3. Digit 0 = O8H
Source Program:
LXI SP, 27FFH : Initialize stack pointer
LDA 6000H : Get the binary number in accumulator
CALL SUBROUTINE : Call subroutine
HLT : Terminate program execution
Subroutine to convert binary number into its equivalent BCD number:
PUSH B : Save BC register pair contents
PUSH D : Save DE register pair contents
MVI B, 64H : Load divisor decimal 100 in B register
MVI C, 0AH : Load divisor decimal 10 in C register
MVI D, 00H : Initialize Digit 1
MVI E, 00H : Initialize Digit 2
STEP1: CMP B : Check if number < Decimal 100
JC STEP 2 : if yes go to step 2
SUB B : Subtract decimal 100
INR E : update quotient
JMP STEP 1 : go to step 1
STEP2: CMP C : Check if number < Decimal 10
JC STEP 3 : if yes go to step 3
SUB C : Subtract decimal 10
INR D : Update quotient
JMP STEP 2 : Continue division by 10
STEP3: STA 6100H : Store Digit 0
MOV A, D : Get Digit 1
STA 6101H : Store Digit 1
MOV A, E : Get Digit 2
STA 6102H : Store Digit 2
POP D : Restore DE register pair
POP B : Restore BC register pair
RET : Return to main program
