8086 Assembly Language Program to print from A to Z

Question

I have the following program that would print from A to Z with a space in between them. In the following program I understood the rest of the code but didn't understand why the PUSH DX and POP DX instructions are used. If I run the code without using PUSH DX and POP DX, it would just print "!" instead of the characters.

.model small
.stack
.data
    VAL DB 'A'

.code
    MAIN PROC
        SPACE MACRO
            MOV DL, ' '
            MOV AH, 02h;
            INT 21H    
        ENDM
        
        MOV AX, @DATA
        MOV DS, AX
        
        MOV CL, 26
        MOV DL, 65 ; MOV DL, VAL
        
        PRINT:
            MOV AH, 02H
            INT 21H
            PUSH DX
            SPACE
            POP DX
            INC DL
            DEC CL
            JNZ PRINT
            
        MOV AH, 4CH
        INT 21H        
            
    MAIN ENDP
    END MAIN

Related: a Codegolf.SE answer: [Print the ASCII Code page](https://codegolf.stackexchange.com/a/249314) and [another answer on the same question](https://codegolf.stackexchange.com/questions/249247/print-the-ascii-code-page/249392#249392) — Peter Cordes, Jul 02 '22 at 02:25

score 5 · Accepted Answer · answered Jan 01 '14 at 12:50

The DX register starts out loaded with 65, the ASCII code for A, the DH and DL registers are the upper and lower half, respectively of DX.

The SPACE macro loads a 32 (the ASCII code for space) into DL, overwriting whatever is there. The PUSH DX POP DX saves and restores the state of the register while the space in-between characters is printed. As to why you see !, that is because without restoring DX after the space has been printed, you will simply increment 32 to 33 and print that character.

Roman Hocke · Answer 2 · 2014-01-01T12:55:22.397

3

In SPACE You set DL to space character, but in main loop You use DL as the A to Z chars. Expansion of SPACE overwrites DL, so You store it to the stack before it and retreive it back after the SPACE.

edited Jan 01 '14 at 12:55

answered Jan 01 '14 at 12:48

Roman Hocke

4,137
1
20
34

score 0 · Answer 3 · edited Apr 12 '18 at 17:47

0

.MODEL SMALL
.STACK 100H
.DATA

MAIN PROC 

    MOV BL,65
    MOV CL,26


PRINT:

    MOV AH,2
    MOV DL,BL
    INC BL
    DEC CL
    INT 21H


    MOV DL,0DH
    INT 21H
    MOV DL,0AH
    INT 21H

 JNZ PRINT

  MOV AH,4CH
  INT 21H

MAIN ENDP
END MAIN

edited Apr 12 '18 at 17:47

Jules Dupont

7,259
7
39
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answered Apr 12 '18 at 16:40

Imran Khan

1

it's a simple way to print capital letter A to Z – Imran Khan Apr 12 '18 at 16:42

score 0 · Answer 4 · answered Jul 01 '22 at 20:24

0

org 100h

.model small
.data 
     msg1 db 13,10, " Print A to Z characters: $"
     
.code 
main proc
    
    mov ax,@data
    mov dx,ax
    
    mov dx,offset msg1
    
    mov ah,9
    int 21h
    
 mov cx,25
mov dx,65
mov ah,2
int 21h
labelname:

 
   inc dx
mov ah,2
int 21h  

loop labelname  
               
  mov ah,4ch
int 21h             


main endp
end main

answered Jul 01 '22 at 20:24

Areeba Hamood

1
1

This has the same (non)segment-register bug as your answer on [How to convert the uppercase to lowercase character in assembly language 8086](https://stackoverflow.com/a/72833971) - a DOS `.com` program (`org 100h`) shouldn't use `@data` at all, leave `DS` unmodified. This happens to work because you write it to DX, not DS, which does nothing: you might as well have done `mov ax, 1` / `mov dx, 2` for all the effect it has on your program. It's a confusing distraction for beginners, code that shouldn't be there but looks almost like a standard idiom for `.exe` programs. – Peter Cordes Jul 02 '22 at 02:30

8086 Assembly Language Program to print from A to Z

4 Answers4