Displaying the even characters of a string in Assembly (MASM)-CodePudding

I want to print out the even charachters of a string. So far I got this code (this code print out the whole string):

.MODEL SMALL
.STACK
.DATA
    adat DB "Test test",0
.CODE
main proc
    MOV AX, DGROUP
    MOV DS, AX
    LEA BX, adat
    new:
        MOV DL, [BX] 
        OR DL, DL 
        CALL WRITE_CHAR
        JZ stop 
        INC BX
        JMP new
    stop: 
        MOV AH,4Ch ;Kilépés 
        INT 21h
main endp

write_char proc                                 
  PUSH  AX                  
  MOV   AH, 2               
  INT   21h                 
  POP   AX                  
  RET                       
write_char endp

END main

So far I have been able to get there. I've tried a few things before, but they didn't work.

CodePudding user response：

OR DL, DL 
CALL WRITE_CHAR
JZ stop

By the time the CALL WRITE_CHAR returns, the flags that were defined from the OR DL, DL instruction will have vanished! So it's possible the loop doesn't run the required number of times.

What are the even characters in a string ?

Is it the characters that reside on an even offset within the string?

main proc
    mov  ax, DGROUP
    mov  ds, ax
    xor  bx, bx       ; (*)
    jmp  char
  next:
    inc  bx           ; The sought for EVEN offset just became ODD here ...
    test bx, 1
    jz   char         ; ... so using opposite condition to skip
    mov  ah, 02h      ; DOS.WriteChar
    int  21h
  char:
    mov  dl, adat[bx] 
    test dl, dl
    jnz  next

    mov  ax, 4C00h    ; DOS.Terminate
    int  21h
main endp

Is it the characters at the 2nd, 4th, 6th, ... place in the string, so at odd offsets?

main proc
    mov  ax, DGROUP
    mov  ds, ax
    xor  bx, bx       ; (*)
    jmp  char
  next:
    inc  bx           ; The sought for ODD offset just became EVEN here ...
    test bx, 1
    jnz  char         ; ... so using opposite condition to skip
    mov  ah, 02h      ; DOS.WriteChar
    int  21h
  char:
    mov  dl, adat[bx]
    test dl, dl
    jnz  next

    mov  ax, 4C00h    ; DOS.Terminate
    int  21h
main endp

Or is it the characters whose ASCII code is an even number?

main proc
    mov  ax, DGROUP
    mov  ds, ax
    xor  bx, bx
    jmp  char
  next:
    inc  bx
    test dl, 1        ; Testing ASCII
    jnz  char         ; Skip on ODD code
    mov  ah, 02h      ; DOS.WriteChar
    int  21h
  char:
    mov  dl, adat[bx] 
    test dl, dl
    jnz  next

    mov  ax, 4C00h    ; DOS.Terminate
    int  21h
main endp

(*) Instead of using an extra register to test for even-ness, we can work from the address register by which we address the string.

In these code snippets, repeating the unconditional jump was avoided by placing the test for string termination near the bottom of the loop and executing a one-time-only jump to it.