Paolo Carlini wrote:
Hi ho, ho!! ;)
It worked with me.
Try a recent gcc (eg, 4.3.x) and you will get the same, actually
expected, result of the original poster.
Paolo.
I believe this is a bug. I agree that -ffast-math will not always comply 100%
with IEEE, as advertised. But, if I
Neal Becker wrote on 31 July 2008 12:42:
Paolo Carlini wrote:
Hi ho, ho!! ;)
It worked with me.
Try a recent gcc (eg, 4.3.x) and you will get the same, actually
expected, result of the original poster.
Paolo.
I believe this is a bug. I agree that -ffast-math will not always comply
Dave Korn wrote on 31 July 2008 12:45:
Neal Becker wrote on 31 July 2008 12:42:
If it is really intended not to work, then at least
documentation should state that.
It does, as was already explained to you; re-read the thread.
Actually I can be more use than that, I'll show you the
Hi Ho!
--- On Tue, 7/29/08, Neal Becker [EMAIL PROTECTED] wrote:
Paolo Carlini wrote:
... ah, ok, now I see what you meant, you meant that x is *not* finite,
still, std::isfinite(x) != 0. Still, testcase badly needed...
Paolo.
#include cmath
#include stdexcept
int main () {
Hi ho, ho!! ;)
It worked with me.
Try a recent gcc (eg, 4.3.x) and you will get the same, actually
expected, result of the original poster.
Paolo.
gcc-4.3.0-8.x86_64
I have test code that does passes std::isfinite (x), yet if I print the
values to std::cout the value printed is 'inf'. Is std::isfinite (x)
broken?
Neal Becker wrote:
gcc-4.3.0-8.x86_64
I have test code that does passes std::isfinite (x), yet if I print the
values to std::cout the value printed is 'inf'. Is std::isfinite (x)
broken?
Whatever bug it may have - it can, of course - std::isfinite returns an
*int*, therefore your statement
Paolo Carlini wrote:
Neal Becker wrote:
gcc-4.3.0-8.x86_64
I have test code that does passes std::isfinite (x), yet if I print the
values to std::cout the value printed is 'inf'. Is std::isfinite (x)
broken?
Whatever bug it may have - it can, of course - std::isfinite returns an
On Mon, Jul 28, 2008 at 1:52 PM, Neal Becker [EMAIL PROTECTED] wrote:
Paolo Carlini wrote:
Neal Becker wrote:
gcc-4.3.0-8.x86_64
I have test code that does passes std::isfinite (x), yet if I print the
values to std::cout the value printed is 'inf'. Is std::isfinite (x)
broken?
Whatever
... ah, ok, now I see what you meant, you meant that x is *not* finite,
still, std::isfinite(x) != 0. Still, testcase badly needed...
Paolo.
Neal Becker wrote:
I found that compiling without -ffast-math would allow std::isfinite to
work. Sorry if the statement was confusing. The code looks something
like:
[calculate x]
if (not isfinite (x))
throw std::runtime_error (blah)
Well, -ffast-math implies -ffinite-math-only, I think
Paolo Carlini wrote:
... ah, ok, now I see what you meant, you meant that x is *not* finite,
still, std::isfinite(x) != 0. Still, testcase badly needed...
Paolo.
#include cmath
#include stdexcept
int main () {
double x = log (0);
if (not std::isfinite (x)) {
throw std::runtime_error
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