asm code and an inout function argument

[Vladimir A. Reznichenko]

Denis Koroskin Wrote:

> On Fri, 15 May 2009 14:41:35 +0400, Vladimir A. Reznichenko <kalessil at gmail.com> wrote:
> 
> > Denis Koroskin Wrote:
> >
> >> On Fri, 15 May 2009 14:24:16 +0400, Vladimir A. Reznichenko  
> >> <kalessil at gmail.com> wrote:
> >>
> >> > I have a function:
> >> >
> >> > void test (inout uint a)
> >> > {
> >> > 	asm
> >> > 	{
> >> > 		mov a, 0x25;
> >> > 	}
> >> > }
> >> >
> >> > The trouble is that the function's call doesn't change the a variable.
> >> > Any ideas?
> >> >
> >>
> >> I believe your code is incorrect. This is how it should be done:
> >>
> >> import std.stdio;
> >>
> >> void test (out uint a)
> >> {
> >>     asm
> >>     {
> >>         mov EDX, a;
> >>         mov [EDX], 0x25;
> >>     }
> >> }
> >>
> >> void main()
> >> {
> >>     uint a = 0;
> >>     test(a);
> >>
> >>     writefln("0x%x", a);
> >> }
> >>
> >> Perhaps, errors like yours could be flagged at compile time? If so, an  
> >> enhancement request would be nice.
> >
> >
> > Thank you, Denis.
> 
> You are wellcome.
> 
> But I stand corrected - your original code was correct, it just didn't do what you expected (I replaced inout with pointer for clarity):
> 
> void test (uint* a)
> {
>     writefln("0x%x", a); // prints 0x12FE88, may differ
>     asm {
>         mov a, 0x25;
>     }
>     writefln("0x%x", a); // prints 0x25, i.e. you were modifying 'a', not '*a'
> }
> 
> The following would be correct, but it is disallowed and silently ignored:
> void test (uint* a)
> {
>     asm {
>         mov [a], 0x25; // no warning is issued, works as if there were no brackets around a, is that correct behavior?
>     }
> }

It looks like "inout/ref uint a" is equal to "uint* a" but the situation when we write D's code "a = 5" means "*a = 5". This is not obvious, at all. So when I wrote asm code, it wouldn't work.

Interesting implementation of inout arguments )
What's more interesting is that it wasn't reflected in inline asm documentation.