Write a programme in 8086 assembly language that accepts two input digits and creates a packed BCD number in AL register?

Answer 1

There should be a jump over the variables/array declaration:

org 100h

jmp code

myArray dw 2, 12, 8, 52, 108

code: mov si, 0

mov ax, myArray[si]

ret

For the computer all bytes look the same, it cannot determine if it's an instruction or a variable. Here is an example of MOV AL, 5 instruction that can be coded with simple variable declarations:

org 100h

byte1 db 176

byte2 db 5

ret

When you run this program in emulator you can see that bytes 176 and 5 are actually assembled into:

MOV AL, 5

This is very typical for Von Neumann Architecture to keep data and instructions in the same memory, It's even possible to write complete program by using only DB (define byte) directive.

org 100h

db 235 ; jump...

db 6 ; 6 - six bytes forward (need to skip characters)

db 72 ; ascii code of 'H'

db 101 ; ascii code of 'e'

db 108 ; ascii code of 'l'

db 108 ; ascii code of 'l'

db 111 ; ascii code of 'o'

db 36 ; ascii code of '$' - DOS function prints until dollar.

db 186 ; mov DX, .... - DX is word = two bytes

db 2 ; 02 - little end

db 1 ; 01 - big end

db 180 ; mov AH, ....

db 9 ; 09

db 205 ; int ...

db 33 ; 21h - 33 is 21h (hexadecimal)

db 195 ; ret - stop the program.

8086 and all other Intel's microprocessors store the least significant byte at a lower address. 102h is the address of 'H' character = org 100h + 2 bytes (jmp instruction). The above assembly code produces identical machine code to this little program:

org 100h

jmp code

msg db 'Hello$'

code: mov DX, offset msg

mov AH, 9

int 21h

ret

If you open the produced ".com" file in any hex editor you can see hexadecimal values, every byte takes two hexadecimal digits, for example 235 = EB, etc... memory window of the emulator shows both hexadecimal and decimal values.