std::isfinite broken?

std::isfinite broken?

[Neal Becker]

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?

Re: std::isfinite broken?

[Paolo Carlini]

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 seems at the very least rather weird. A 
self-contained testcase is badly needed.

Paolo.

Re: std::isfinite broken?

[Neal Becker]

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
> *int*, therefore your statement seems at the very least rather weird. A
> self-contained testcase is badly needed.
> 
> Paolo.

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")

Re: std::isfinite broken?

[Richard Guenther]

On Mon, Jul 28, 2008 at 1:52 PM, Neal Becker <ndbecker2@gmail.com> 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 bug it may have - it can, of course - std::isfinite returns an
>> *int*, therefore your statement seems at the very least rather weird. A
>> self-contained testcase is badly needed.
>>
>> Paolo.
>
> I found that compiling without -ffast-math would allow std::isfinite to
> work.

That's by desing.  With -ffast-math you are assessing that Infs and NaNs
do not occur, so the compiler optimizes the call to isfinite.

Richard.

Re: std::isfinite broken?

[Paolo Carlini]

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 you have to pass 
-fno-finite-math-only if you want to deal correctly with infinities.

Paolo.

Re: std::isfinite broken?

[Paolo Carlini]

... 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.

Re: std::isfinite broken?

[Neal Becker]

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 ("not finite");
  }
}

Compiled with -O3 -ffast-math will not throw.

Re: std::isfinite broken?

[Eus]

Hi Ho!

--- On Tue, 7/29/08, "Neal Becker" <ndbecker2@gmail.com> 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 () {
>   double x = log (0);
>   if (not std::isfinite (x)) {
>     throw std::runtime_error ("not finite");
>   }
> }
>
> Compiled with -O3 -ffast-math will not throw.

It worked with me.

Here is the record:

eus@experimental:~/shared/del/PHP$ cat x.c

#include <iostream>
#include <cmath>
#include <stdexcept>

int main () {
  double x = log (0);
  if (not std::isfinite (x)) {
	std::cout << "Inside if" << std::endl;
  }

  std::cout << "Outside if" << std::endl;
  return 0;
}
eus@experimental:~/shared/del/PHP$ g++ -o a -O3 -ffast-math x.c
eus@experimental:~/shared/del/PHP$ ./a
Inside if
Outside if

eus@experimental:~/shared/del/PHP$ g++ -v
Reading specs from /usr/lib/gcc-lib/i486-slackware-linux/3.3.6/specs
Configured with: ../gcc-3.3.6/configure --prefix=/usr --enable-shared --enable-threads=posix --enable-__cxa_atexit --disable-checking --with-gnu-ld --verbose --target=i486-slackware-linux --host=i486-slackware-linux
Thread model: posix
gcc version 3.3.6

Best regards,
Eus


      

Re: std::isfinite broken?

[Paolo Carlini]

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.

Re: std::isfinite broken?

[Neal Becker]

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 explicity ask isfinite(), I expect it to work.  If it is really intended not to work, then at least documentation should state that.

RE: std::isfinite broken?

[Dave Korn]

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
> 100% with IEEE, as advertised.  But, if I explicity ask isfinite(), I
> expect it to work.  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.


    cheers,
      DaveK
-- 
Can't think of a witty .sigline today....

RE: std::isfinite broken?

[Dave Korn]

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 relevant bits of
the manual:

-ffast-math'
    Sets `-fno-math-errno', `-funsafe-math-optimizations',
    `-fno-trapping-math', `-ffinite-math-only' and
                           ^^^^^^^^^^^^^^^^^^^
    `-fno-signaling-nans'.

`-ffinite-math-only'
     Allow optimizations for floating-point arithmetic that assume that
     arguments and results are not NaNs or +-Infs.


  So, to address your point:

>>But, if I explicity ask isfinite(), I expect it to work.  


  Fine.  But what do you expect to happen if you explicitly tell it that no
number in your program will ever be infinite, and then call isinfinite?  You
told GCC that it could assume isinfinite will always return zero, so it took
your word for that and replaced all the calls by a constant zero.  What do
you want to happen in those circumstances?




    cheers,
      DaveK
-- 
Can't think of a witty .sigline today....