q-lang-users

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[John C. <co...@cc...>]

Abdulaziz Ghuloum scripsit:

> The test [for infinity] is obviously bogus since f==f+1.0 for large values of f.
> (try 1e16)

Oops, quite right.

The expression "f != 0.0 && f == f + f" will work, I think.
0.0 and NaN are rejected by the left side of "&&", and all
other finite values by the right side.

-- 
Andrew Watt on Microsoft:                       John Cowan
Never in the field of human computing           co...@cc...
has so much been paid by so many                http://www.ccil.org/~cowan
to so few! (pace Winston Churchill)

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[John C. <co...@cc...>]

Abdulaziz Ghuloum scripsit:

> Out of curiosity, why not simply do "f==1.0/0.0 || f==-1.0/0.0" to test
> if f is +inf.0/-inf.0?

Umm.  Too obvious, I guess.  I started working these out in
Chicken, where dividing by zero throws an exception, so I
wanted some other method.  But in C, of course, there's no
such issue.

-- 
One art / There is                      John Cowan <co...@cc...>
No less / No more                       http://www.ccil.org/~cowan
All things / To do
With sparks / Galore                     -- Douglas Hofstadter

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[Thomas C. <ch...@we...>]

On Wed, 21 Jun 2006, John Cowan wrote:

> Abdulaziz Ghuloum scripsit:
>
>> Out of curiosity, why not simply do "f==1.0/0.0 || f==-1.0/0.0" to test
>> if f is +inf.0/-inf.0?
>
> Umm.  Too obvious, I guess.  I started working these out in
> Chicken, where dividing by zero throws an exception, so I
> wanted some other method.  But in C, of course, there's no
> such issue. [...]

Hello,

are you sure this works on every architecture? I dimly recall having seen 
systems where NaN != NaN holds, so you have to use the isnan predicate 
function from the C math library...

cu,
Thomas

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[Albert G. <Dr....@t-...>]

John Cowan wrote:
> Abdulaziz Ghuloum scripsit:
> 
>>Out of curiosity, why not simply do "f==1.0/0.0 || f==-1.0/0.0" to test
>>if f is +inf.0/-inf.0?
> 
> Umm.  Too obvious, I guess.  I started working these out in
> Chicken, where dividing by zero throws an exception, so I
> wanted some other method.  But in C, of course, there's no
> such issue.

But you want the code at least not to do any harm on systems with _no_ 
IEEE floats. I'm not sure, but couldn't dividing by 0.0 on such systems 
kill your program with SIGFPE?

Another point is that division is a much more costly operation than 
adding and subtracting.

BTW, should I cc the Chicken list in this thread?

-- 
Dr. Albert Gr"af
Dept. of Music-Informatics, University of Mainz, Germany
Email:  Dr....@t-..., ag...@mu...
WWW:    http://www.musikinformatik.uni-mainz.de/ag

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[John C. <co...@cc...>]

Albert Graef scripsit:

> But you want the code at least not to do any harm on systems with _no_ 
> IEEE floats. I'm not sure, but couldn't dividing by 0.0 on such systems 
> kill your program with SIGFPE?

I don't know.  I've never done serious floating-point work that
had to run on a non-IEEE system (everything I did on a PDP-8, PDP-11,
or pre-IEEE VAX was pretty much either integer-only or floats-as-integers,
as in most Basics).

> Another point is that division is a much more costly operation than 
> adding and subtracting.

True.

> BTW, should I cc the Chicken list in this thread?

Might as well, though your postings might or might not go through.

-- 
That you can cover for the plentiful            John Cowan
and often gaping errors, misconstruals,         http://www.ccil.org/~cowan
and disinformation in your posts                co...@cc...
through sheer volume -- that is another
misconception.  --Mike to Peter

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[Albert G. <Dr....@t-...>]

John Cowan wrote:
> The expression "f != 0.0 && f == f + f" will work, I think.
> 0.0 and NaN are rejected by the left side of "&&", and all
> other finite values by the right side.

Ok, I'll fix that, thanks.

-- 
Dr. Albert Gr"af
Dept. of Music-Informatics, University of Mainz, Germany
Email:  Dr....@t-..., ag...@mu...
WWW:    http://www.musikinformatik.uni-mainz.de/ag

Re: [q-lang-users] [Chicken-users] Re: IEEE float arithmetic

[Albert G. <Dr....@t-...>]

John Cowan wrote:
> The expression "f != 0.0 && f == f + f" will work, I think.
> 0.0 and NaN are rejected by the left side of "&&", and all
> other finite values by the right side.

Thinking about it some more I think that "!isnan(f) && isnan(f-f)" 
(given John's definition of isnan()) might be an even better way to do 
this. f-f can never be nan on an IEEE float system unless f is either 
nan or inf. And with non-IEEE floats this test should always fail 
because (presumably) f==f will always be true and hence isnan will 
always be false. Right?

This test only relies on the algebraic properties of inf and nan and 
should be about the most efficient way to do it, too, as it just needs a 
single subtraction and two comparisons.

-- 
Dr. Albert Gr"af
Dept. of Music-Informatics, University of Mainz, Germany
Email:  Dr....@t-..., ag...@mu...
WWW:    http://www.musikinformatik.uni-mainz.de/ag