The requested program should be composed of three parts:

1. set up pointers to data
2. display data
3. exit

Actually all the following programs DON'T include part 3, but read on. The data (command line) can be printed either as a single string, or character by character.

APPROACH 1: Print single string

For the first case, part 2 would be:

        mov     ah, 9
mov     dx, 81h
int     21h

that makes 7 bytes, leaving only 4 bytes to replace the last CR with a '$', which are too few. (Actually, if the user would type a $ as the last character in the comand line, this would make the smallest possible program.) The short- est program I managed to write is:

        shr     si,1                    ; D1 EE
lodsb                           ; AC
push    si                      ; 56
add     si,ax                   ; 03 F0
mov     byte ptr [si],'$'       ; C6 04 24
xcgh    bp,ax                   ; 95
pop     dx                      ; 5A
int     21                      ; CD 21

For the second case, the smallest program would be this:

        ; Solution I
mov     dx, 81h                 ; BA 81 00
mov     cl, ds:[80h]            ; 8A 0E 80 00
mov     ah, 40h                 ; B4 40
int     21h                     ; CD 21

The first two lines are part 1 (set up pointers) and the other two are part 2 (display string). If you think that something is missing you're right: we don't set BX (the handle).

APPROACH 2: Print char by char

Of course the second interrupt is better, since there is no need to load ah with a function value. In addition, INT 29 reads the char from AL, so it can be used together with LODSB.

The first way to implement this approach is to minimize part 2 (display loop). A program that does this is the following:

        ; Solution II
mov     si, 80h ; BE 80 00
lodsb           ; AC
mov     cl, al  ; 8A C8
more:   lodsb           ; AC
int     29h     ; CD 29
loop    more    ; E2 FB

This program printed CX characters. The second way to print the string is to print up to the CR. Here is how:

        ; Solution III
mov     si, 81h         ; BE 81 00
more:   lodsb                   ; AC
int     29h             ; CD 29
cmp     al, 13          ; 3C 0D
jne     more            ; 75 F9
nop                     ; 90

Yes, the last instruction IS a NOP. So we have an 11-byte program that works, and even has a NOP in it. Removing the NOP creates an even crazier program that is 10 bytes long, displays it's command line AND waits for a key press before terminating... Actually solution II, by substituting MOV SI,80h with SHR SI,1, does the same thing (10 bytes that display the command line and wait for the user to press a key).

BTW: I really don't know why these programs work, though I have one or two theories...

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