conversion of BCD to HEX or BCD to BINARY or DECIMAL to HEX or DECIMAL to BINARY

Write an assembly language program in a 8085 microprocessor for the conversion of BCD
to HEX or BCD to BINARY or DECIMAL to HEX or DECIMAL to BINARY.

INPUT: (25)10
OUTPUT: 19H

NOTE : calculation done behind the scene is in binary & but in memory it is show us in hexadecimal

Approsch : 

Certainly! Here's a simplified explanation of the approach used in the assembly code for converting BCD to HEX:

1 ) Initialization:
Set up the memory address where the BCD number is stored.
Load the first byte of the BCD number into the accumulator register.
2 ) Extracting the Left 4 Bits: 
Mask the lower 4 bits to get only the left 4 bits of the BCD number. => using AND with FOh
Since the left 4 bits are in the most significant nibble but in unused form, shift them to the least significant nibble.
3 ) Multiplying by 10:
  Way 1 : Use a loop to multiply the left 4 bits by 10 (since each BCD digit is multiplied by 10 when converting to hexadecimal).
  Way 2 : Wihtout loop
4 ) Extracting the Right 4 Bits:
  Move the accumulator to another register (D register) to preserve the left 4 bits.
  Mask the original BCD number to get only the right 4 bits.
5 ) Adding the Right 4 Bits:
Add the right 4 bits to the previously multiplied left 4 bits.
6 ) Storing the Result:
Store the final result (the converted HEX value) in a specified memory address. HLT the program

Dry Run : 

25 => 0010 0101 
extract left 4 bits : 25 & f0 => 20
                      20>>4 => 2
multiply with 10 : 2*10=20 => 0001 0100 (in bianry ) => 14(in memory because it is show us in hexadecimal)
extract right 4 bits : 25 & 0f => 05
add : multiplied bit + ectracted left 4 bits
      14 + 05 = 19h => answer


Code :

lxi h,2000h
;mvi h,25h

mov a,m
ani f0h; step 1 : fetch left 4 bits

;step 2 : it is in unused form because we need only left 4 bits to multiply 10 
;but we get 8 bits with 4 bits as most left 4 bits 
;we need to convert into least left 4 bits
;bring into used form right shift 4 times
rrc
rrc
rrc
rrc
; step 3 : way -1 : multiply with 10 using reapeted addition using lopp
mvi c,09h
mov b,a
loop:add b
     dcr c
     jnz loop

way - 2 : repeated addition without loop 
;add a
;mov d,a
;add a
;add a
;add d

; step 4 : fetch right 4 bits
mov d,a
mov a,m
ani 0fh

; step 5 : add fetched right 4 bits to multipled bits
add d
sta 2010h
mov d,a
hlt