Dos tidbit: register-based Watcall parameter order

It has occurred to me that in the course of writing DOS programs (some of which even work!), I spend an awful lot of time searching through documentation.  Which is, you know...fine.  It's part of the charm of retro coding.

However, this ancient source has an inaccuracy which has bitten me twice (the second time because I forget about 99% of what I read).  Not that I bear the author any ill-will - if a document as detailed and comprehensive as that contained no inaccuracies at all, I would be worshiping the author as some kind of god.  It might even have been accurate at the time of writing (2010CE).  There's plenty more in that site which I intend to dig through and learn from as well.

Anyway.  What bit me:
When using register-based watcall* to call an asm procedure from C/C++, 16-bit parameters are passed in this order: AX, DX, BX, CX. 

*in OpenWatcom V2 at time of writing.

As I read it, the document says it is in the order AX, BX, CX, DX.  This caused me MUCH confusion and my fledgling asm-debugging skills took some hours to dig out the cause of my crashes.

A useful by-product of my toil was that I now have somewhat-tested asm routines which are essentially a debug-print statement.  OutputHex writes the ax register value to the DOS screen as hex:
 

OutputHexDigit PROC WATCOM_C public   
        and ax,0xf
        cmp ax, 0xa
        jb normalDigit
        add ax, 0x37
        jmp outputDigit
       
    normalDigit:
        add ax, 0x30
   
    outputDigit:
        mov ah, 0Eh
        int 10h
    ret
OutputHexDigit ENDP

OutputHex PROC WATCOM_C public   
    push ax
    shr ax,12
    and ax,0xf
    call OutputHexDigit
   
    pop ax
    push ax
    shr ax,8
    and ax,0xf
    call OutputHexDigit

    pop ax
    push ax
    shr ax,4
    and ax,0xf
    call OutputHexDigit

    pop ax
    and ax,0xf
    call OutputHexDigit
   
    mov al,13
    int 10h
   
    mov al,10
    int 10h
   
    ret
OutputHex ENDP