On Monday 02 October 2006 20:53, Gabriel Dos Reis wrote:
[...]
>
> | the result of the "same" multiplication considered as complex *
> | complex (vs complex * real) has a different sign for the zero
> | imaginary component.
>
> Thanks for the example. I'm not sure this was anticipated by the C++
>
On Tuesday 03 October 2006 11:08, Joseph S. Myers wrote:
> On Tue, 3 Oct 2006, Jan van Dijk wrote:
> > More serious is the fact that the compiler translates 1*(Inf,Inf) into
> > (NaN,NaN). This is plain wrong, but, as Joseph mentioned, the solution
> > requires the implementation of mixed-mode MULT
On Tue, 3 Oct 2006, Jan van Dijk wrote:
> More serious is the fact that the compiler translates 1*(Inf,Inf) into
> (NaN,NaN). This is plain wrong, but, as Joseph mentioned, the solution
> requires the implementation of mixed-mode MULT_EXPRESSIONS, since apparently
>
> 1*(Inf,Inf) == (Inf
On Monday 02 October 2006 19:39, Gabriel Dos Reis wrote:
> "Joseph S. Myers" <[EMAIL PROTECTED]> writes:
> | On Mon, 2 Oct 2006, Jan van Dijk wrote:
> | > On Monday 02 October 2006 12:57, Joseph S. Myers wrote:
> | > > On Mon, 2 Oct 2006, Jan van Dijk wrote:
[...]
> | > Triggered by 1*(Inf,0) = (In
On Mon, 2 Oct 2006, Paolo Carlini wrote:
> I'm not sure if the following is exactly Joseph' point, but I'd like to know
> your opinion about it anyway: if you look to Comment #19 in the audit trail of
> PR 28408, I noticed that, due to the rule about signed zero (with default
> rounding):
> (
Paolo Carlini <[EMAIL PROTECTED]> writes:
| Hi Gaby
|
| > | > My question was a slightly different one. To me it is not clear
| > whether the | > standard allows the treatment of (r,0) as r in
| > complex operations. For | > example: is it allowed to handle
| > (r,0)*(x,y) as r*(x,y)?
| > | | I d
Hi Gaby
| > My question was a slightly different one. To me it is not clear whether the
| > standard allows the treatment of (r,0) as r in complex operations. For
| > example: is it allowed to handle (r,0)*(x,y) as r*(x,y)?
|
| I don't think so; at least, it might affect negative 0.
Hmm, how
"Joseph S. Myers" <[EMAIL PROTECTED]> writes:
| On Mon, 2 Oct 2006, Jan van Dijk wrote:
|
| > On Monday 02 October 2006 12:57, Joseph S. Myers wrote:
| > > On Mon, 2 Oct 2006, Jan van Dijk wrote:
| > > > * the C99 and C++ standards say *nothing* about the details of compex
| > > > multiplication
On Mon, 2 Oct 2006, Jan van Dijk wrote:
> On Monday 02 October 2006 12:57, Joseph S. Myers wrote:
> > On Mon, 2 Oct 2006, Jan van Dijk wrote:
> > > * the C99 and C++ standards say *nothing* about the details of compex
> > > multiplication
> >
> > The C99 standard says that real operands aren't co
On Monday 02 October 2006 12:57, Joseph S. Myers wrote:
> On Mon, 2 Oct 2006, Jan van Dijk wrote:
> > * the C99 and C++ standards say *nothing* about the details of compex
> > multiplication
>
> The C99 standard says that real operands aren't converted to complex, but
> as I note in bug 24581, the
On Mon, 2 Oct 2006, Jan van Dijk wrote:
> * the C99 and C++ standards say *nothing* about the details of compex
> multiplication
The C99 standard says that real operands aren't converted to complex, but
as I note in bug 24581, the compiler doesn't expect PLUS_EXPR and
MULT_EXPR to have argume
Hi Jan,
Hello,
I was recently bitten by gcc's handling of complex multiplication. My program
is in C++, but since std::complex uses C99's complex types, my questions
below apply to C as well.
I only want to say more explicitely that apparently both C and C++ are
affected, and in *i
Hello,
I was recently bitten by gcc's handling of complex multiplication. My program
is in C++, but since std::complex uses C99's complex types, my questions
below apply to C as well.
In my program, I have r*c, where r is an fp type and c a complex. In my
calculation I sometimes encoun
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