'Assembly 16 bit illogical mistake (?)

I don't understand why this code print me the char I've insert back and the alphabet string if nobody told him to do it...

.model
.stack 100h
.data

lettere DB 10 DUP(?)
frase1 DB 10,13,"inserisci 10 caratteri$"
frase2 DB ">>$"
caporiga DB 10,13,"$"
counter DB 0
char DB ?

alpha DB "abcdefghijklmnopqrstuvwxyz$"

.code
.startup

  lea dx,frase1
  mov ah,09h
  int 21h

  lea dx,caporiga
  mov ah,09h
  int 21h

insertLoop:

  inc counter

  lea dx,counter
  mov ah,02h
  int 21h

  lea dx,frase2
  mov ah,09h
  int 21h

  mov ah,01h
  int 21h
  mov lettere+(counter-1),al

  lea dx,caporiga
  mov ah,09h
  int 21h

  cmp counter,10
  JB insertLoop

  mov ah,4ch
  int 21h
end
assemblyx86-1616-bit


Solution 1:[1]

I don't understand why this code print me the char I've insert back and the alphabet string if nobody told him to do it...

The mov lettere+(counter-1),al instruction does not do what you want.
MASM will add the (offset)addresses of lettere (0) and counter (41), and then subtract 1, yielding the address 40 where the instruction then overwrites an important $ string terminator. The next time that your program tries to print a newline caporiga, the DOS.PrintString function 09h will continue printing characters until it finds the $ that ends your alphabet string.

lea dx,counter
mov ah,02h
int 21h

This is another error. The DOS.PrintCharacter function 02h expects to find a character in the DL register, but this lea dx,counter instruction clearly loads an address.

mov  dl, counter
add  dl, '0'
mov  ah, 02h
int  21h

The correct code first retrieves the count [1,10] from memory and then adds '0' (48) in order to convert into a printable character. This will of course fail when your count is 10. Look in Displaying numbers with DOS to learn how to overcome this problem.

One possible solution:

...

  xor BX, BX        ; Reset counter
insertLoop:
  inc BX            ; Increment counter
  lea dx, [BX + 48] ; -> DL = "1", "2", ...
  mov ah, 02h       ; Won't work when counter is 10
  int 21h

  lea dx, frase2
  mov ah, 09h
  int 21h

  mov ah, 01h
  int 21h
  mov [lettere + BX - 1], al

  lea dx, caporiga
  mov ah, 09h
  int 21h

  cmp BX, 9
  jb  insertLoop

...

Sources

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Source: Stack Overflow

Solution Source
Solution 1 Sep Roland

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