Passing a 1 byte argument to a function?

Question

I want to create a function that receives a 1 byte argument. But I am reading that in x86 I can only push 2 or 4 bytes onto the stack. So should I expect a 2 bytes argument to be passed to my function and then extract my 1 byte? Is this is how to pass a 1 byte argument to my function:

push WORD 123

Pass it as an 8-bit register, or push 16-bit register on the stack (having 8 lower or highest bits as the value you want) — , Dec 30 '14 at 05:44

score 2 · Accepted Answer · answered Dec 30 '14 at 05:44

The stack must be aligned. If you make 16 bit real mode program, the stack must be aligned on 16 bits. If you make 32 bit, protected mode program the stack must be aligned on 32 bits.

But you don't need to pass exactly 1 byte to the function. Just push 16/32 bits and use only the lowest 8 of them in the function. Something like this:

use32
proc  MyFunc, .arg32, .arg16, .arg8
begin
       mov  eax, [.arg32]
       mov  bx, word [.arg16]
       mov  cl, byte [.arg8]
       ret
endp

Main:
       push  ecx   ; CL is arg8
       push  ebx   ; BX is arg16
       push  eax   ; EAX is arg32
       call  MyFunc

; Or shortly:
       stdcall MyFunc, eax, ebx, ecx

Passing a 1 byte argument to a function?

1 Answers1