Re: c/9762: Address of 'char' is incorrect.
On Thu, Feb 20, 2003 at 06:32:19PM -0000, Stephen Kennedy wrote:
>
> OK, the C standard does not say that this should work, so you
> can consider this bug closed.
>
> However, given knowledge of the calling convention of a
> particular machine, you can do neat things such as dynamic
> function binding. See www.drizzle.com/~scottb/gdc/fubi-paper.htm
> for instance.
>
> I've since changed to using assembly, but why does gcc
> return the address of a temp when 'a' is a char and not
> when 'a' is an int?
Because it's passed as an int; you have to convert it to a char on
arrival in the function.
>
> Surprised, not unhappy,
> Stephen.
>
> ---
> Stephen Kennedy <Stephen.Kennedy@havok.com>
> t: +353 1 6693679 f: +353 1 6767094
> Game Developer Frontline Award Winner
> http://www.havok.com/news/release.html
>
> > > In the example below, '&a' is the address of a local
> > copy of 'a' not of 'a'.
> > > if the type of 'a' is changed to int, it works as expected.
> >
> > Works as who expected? Where is the bug? Please quote which part of
> > the C standard is violated. You got an address, why are you unhappy?
> >
> > Neil.
> >
> > > #define TA char
> > > #define TB int
> > > #define TC int
> > >
> > > void foobar(TA a, TB b, TC c);
> > >
> > > int main()
> > > {
> > > foobar(1,2,3);
> > > return 0;
> > > }
> > >
> > > void foobar(TA a, TB b, TC c)
> > > {
> > > printf("a == %i claims %x\n", a, &a);
> > > printf("a == %i really %x\n", (&b)[-1], (&b)-1);
> > > printf("b == %i %x\n", b, &b);
> > > printf("c == %i %x\n", c, &c);
> > > }
> >
>
>
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>
--
Daniel Jacobowitz
MontaVista Software Debian GNU/Linux Developer
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