Re: asm code and an inout function argument
On Fri, 15 May 2009 19:11:55 +0400, Jesse Phillips wrote: > On Fri, 15 May 2009 07:14:50 -0400, Vladimir A. Reznichenko wrote: > >> 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. > > Isn't that the point of a reference, that you don't have to dereference > it? In fact I believe "*a = 5" would be an error when using references. Isn't it an error right now?
Re: asm code and an inout function argument
On Fri, 15 May 2009 07:14:50 -0400, Vladimir A. Reznichenko wrote: > 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. Isn't that the point of a reference, that you don't have to dereference it? In fact I believe "*a = 5" would be an error when using references.
Re: asm code and an inout function argument
Denis Koroskin Wrote:
> On Fri, 15 May 2009 14:41:35 +0400, Vladimir A. Reznichenko
> wrote:
>
> > Denis Koroskin Wrote:
> >
> >> On Fri, 15 May 2009 14:24:16 +0400, Vladimir A. Reznichenko
> >> 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.
Re: asm code and an inout function argument
On Fri, 15 May 2009 14:41:35 +0400, Vladimir A. Reznichenko
wrote:
> Denis Koroskin Wrote:
>
>> On Fri, 15 May 2009 14:24:16 +0400, Vladimir A. Reznichenko
>> 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?
}
}
Re: asm code and an inout function argument
Tim Matthews Wrote:
> On Fri, 15 May 2009 22:29:07 +1200, Robert Fraser
> wrote:
>
> > Vladimir A. Reznichenko 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?
> >
> > Inout variables are pointers.
>
>
> Why is it 'inout' and not 'ref' ?
Aren't they the same?
Re: asm code and an inout function argument
Denis Koroskin Wrote:
> On Fri, 15 May 2009 14:24:16 +0400, Vladimir A. Reznichenko
> 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.
Re: asm code and an inout function argument
On Fri, 15 May 2009 14:24:16 +0400, Vladimir A. Reznichenko
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.
Re: asm code and an inout function argument
On Fri, 15 May 2009 22:29:07 +1200, Robert Fraser
wrote:
Vladimir A. Reznichenko 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?
Inout variables are pointers.
Why is it 'inout' and not 'ref' ?
Re: asm code and an inout function argument
Vladimir A. Reznichenko 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?
Inout variables are pointers.
