Not a number problem

[istillshine]

In my code I used NAN and isnan(x). I found they were convenient to
use. I also noticed that
older C standard does not support NAN and isnan(x).

When I compiled my program using:

gcc -Wall -c

it was fine.


But when I compiled my program using:

gcc -Wall -ansi -pedantic -c

it reported some errors, reporting NAN and isnan(x) not supported.

I knew the options -ansi and -pedantic make things conform to the
older C standard (C90?).

My question are:

1. Is it fine to compile my program using "gcc -Wall -c" instead of
using the more conservative "gcc -Wall -ansi -pedantic -c"?

2. Dose including NAN and isnan(x) hurt the portability of a program,
given high version of gcc is available in both Linux and Windows
(MinGW)?

 

[Nick Keighley]

In my code I used NAN and isnan(x).  I found they were convenient to
use.  I also noticed that
older C standard does not support NAN and isnan(x).

only C99 supports IEEE floating point, and even then
only if a particular macro is defined (

 

[Nick Keighley]

oops! last post sent in error

In my code I used NAN and isnan(x).  I found they were convenient to
use.  I also noticed that
older C standard does not support NAN and isnan(x).

as I said... only c99 offers IEEE support and even then it
is optional.

When I compiled my program using:

gcc -Wall -c

it was fine.

But when I compiled my program using:

gcc -Wall -ansi -pedantic -c

it reported some errors, reporting NAN and isnan(x) not supported.

as they aren't part of the standard

I knew the options -ansi and -pedantic make things conform to the
older C standard (C90?).

probably c90

My question are:

1. Is it fine to compile my program using  "gcc -Wall -c" instead of
using the more conservative "gcc -Wall -ansi -pedantic -c"?

only you can answer this. What is more important to you, IEEE
support or maximal portability?

2. Dose including NAN and isnan(x) hurt the portability of a program,
yes

given high version of gcc is available in both Linux and Windows
(MinGW)?

where is your software likely to run? If you going to
do lots of floating point you may decide to only support
systems that have IEEE support (a lot these days).

You might try putting gcc into its (almost) C99 mode.

The world is NOT confined to Linux and Windows

 

[vippstar]

only C99 supports IEEE floating point, and even then
only if a particular macro is defined (

Hmm.. something went wrong here?
The macro is __STDC_IEC_559__
<snip>

 

[Philip Potter]

oops! last post sent in error



as I said... only c99 offers IEEE support and even then it
is optional.

This is true but misleading. C99 does only offer optional IEEE support,
but the isnan() macro is not part of the IEEE extension.

See n1256 7.12.3.4 - isnan() is defined in math.h in all C99
implementations. (more below)

as they aren't part of the standard

NaN values and the isnan() macro are part of the standard. To be
precise, the standard allows but does not require NaNs to exist. If NaNs
don't exist the isnan() macro must still be provided and always returns
zero. The NAN macro is defined if and only if quiet NaN values are
supported by the float type.

probably c90

Yes, -ansi corresponds to C90. (AFAIK it is equivalent to -std=c90)

only you can answer this. What is more important to you, IEEE
support or maximal portability?

Again, there are more NaN-supporting systems than just IEEE.

yes

Probably.

 

[istillshine]

Then I wonder how people deal with NAN or INFINITY numbers in a
portable manner? I guess this situation arises frequently in
developing numerical softwares. In my program, I need to do something
like the following:


while (1) {
...
x = some expression; /* may cause problem */
if (x is neither NAN nor INFINITY) { /* make sure x is a useful
number */
break;
} else {
/* deal with the problem */
}
}

 

[Walter Roberson]

Then I wonder how people deal with NAN or INFINITY numbers in a
portable manner? I guess this situation arises frequently in
developing numerical softwares.

NaN and Infinity do not occur in all floating point systems.
For example it was not uncommon for single precision floating point
numbers to be in some native format that used the entire value
space instead of reserving some of the potential value space as
special values such as infinity or the various types of NaN.

Thus, "dealing with NaN or infinity numbers in a portable manner"
is dubious -- unless, that is, you choose to restrict your
portability to systems that have defined support for them built in.


A classic test for a NaN, by the way, is

(!(x==x) && !(x!=x))

That is, a NaN does not compare equal to itself, and also does not
compare -not equal- to itself. And hope that the compiler doesn't
optimize the test away thinking that "of course x==x" . A compiler
that doesn't know that NaN's exist might make that mistake, so code
carefully.


If you are working with NaN and infinity, then chances are that you
are also concerned with rounding modes; the manipulation of rounding
modes is usually fairly system-specific before C99 (and
from what I remember, C99 does not capture the full flavour of
rounding modes.)

 

[jacob navia]

Then I wonder how people deal with NAN or INFINITY numbers in a
portable manner? I guess this situation arises frequently in
developing numerical softwares. In my program, I need to do something
like the following:


while (1) {
...
x = some expression; /* may cause problem */
if (x is neither NAN nor INFINITY) { /* make sure x is a useful
number */
break;
} else {
/* deal with the problem */
}
}

Just use isnan() and be done with it. It is standard C.
And if you find a compiler that doesn't support isnan()
throw it away and get a better one.

P.S.
I think

int isnan(double x)
{
if (x != x)
return 1;
return 0;
}

 

[Walter Roberson]

P.S.
I think

int isnan(double x)
{
if (x != x)
return 1;
return 0;
}

Though what happens if x is a signalling NaN? Signalling NaNs trigger
when their value is used. Just testing isnan(x) is not a use of the
value (it's test of the properties), but x != x would be a use of
the value and so would (if I understand correctly) trigger the signal.

 

[christian.bau]

A classic test for a NaN, by the way, is

  (!(x==x) && !(x!=x))

That is, a NaN does not compare equal to itself, and also does not
compare -not equal- to itself.

NaN _always_ compares not equal to itself, that is x != x is _always_
true for NaNs. And (x != y) will _always_ give the same result as ! (x
== y), even with one or more NaNs and/or Infinities involved.

 

[jacob navia]

Though what happens if x is a signalling NaN? Signalling NaNs trigger
when their value is used. Just testing isnan(x) is not a use of the
value (it's test of the properties), but x != x would be a use of
the value and so would (if I understand correctly) trigger the signal.

The signal would be triggered anyway. It is up to you to
hide that signal using the fesetenv() (if you use C99) or
whatever. In ANY case, if you have a signaling NAN as a
result of a computation and the processor has that signal unmasked
your program will crash anyway.

I do not see the point of your objection.

 

[Keith Thompson]

Why not this?

int isnan(double x)
{
return x != x;
}

(Just a minor style point.)

(Either way, the equivalent of isnan() is easy enough to implement
even in C90 that there's no point in rejecting a non-C99 compiler just
for this reason.)

Though what happens if x is a signalling NaN? Signalling NaNs trigger
when their value is used. Just testing isnan(x) is not a use of the
value (it's test of the properties), but x != x would be a use of
the value and so would (if I understand correctly) trigger the signal.

C99 explicitly doesn't define the behavior of signaling NaNs, even in
the optional Annex F (IEC 60559 floating-point arithmetic). The
isnan() function presumably tests for a quiet NaN. As jacob points
out, invoking ``isnan(x)'' if x is a signalling NaN has to evalute x
first, which could result in a trap.

In effect, signalling NaNs are just trap representations.

Note that the language could have provided, and an implementation can
provide, a function that tests for signalling NaNs:

int is_signalling_nan(double *x);

It could, for example, convert the double* argument to unsigned char*
and examine the representation, using system-specific knowledge of
what a signalling NaN looks like. As long as it doesn't access the
value *as floating-point*, it won't invoke UB. (Either this would
have to be a macro, probably using ``sizeof *x'' to determine what
type it points to, or there would have to be versions for float,
double, and long double.)

 

[istillshine]

I got the following message when I compiled my program using "gcc -
Wall -c -ansi -pedantic"

tr.c:1246: warning: implicit declaration of function `isnan'

min.o(.text+0x36d):minimize.c: undefined reference to `isinf'

It seemed isnan() is not there by default. And, how to determine if a
number is infinite (negative or positive)?

Will the following function work?

int isinf(double x)
{
return x == -DBL_MAX || x == DBL_MAX;
}

 

[istillshine]

I just tried the isinf() function I wrote previously, it did not
work. However, I found somewhere in clc an alternative:

static int isinf(double x)
{
return (x == x) && (x != 0) && (x + 1 == x);
}

And tested it using isinf(log(0) and isinf(1.0/0.0). It seemed
working.

 

[Richard Tobin]

return (x == x) && (x != 0) && (x + 1 == x);

I have serious doubts that this will work correctly for large but
non-infinite numbers. For example, consider 1.0e+300. This fits
in an IEEE double. It compares equal to itself. It does not compare
equal to zero. And 1.0e+300 is going to compare equal to 1.0e+300
+ 1 because you don't have nearly enough mantissa bits in an IEEE
double to distinguish the two.[/QUOTE]

Perhaps x == x/2 would work better. You would need the x != 0 in that
case.

-- Richard

 

[istillshine]

It is very good for me to learn this point. Thanks.

 

[istillshine]

Collecting ideas from you, I wrote a header file to handle the isnan
and isinf problem.

#ifndef _NAN_H
#define _NAN_H

#ifndef isnan
#define isnan(x) ((x) != (x))
#endif

#ifndef isinf
#define isinf(x) (((x) == (x)) && ((x) != 0) && ((x)/2 == (x)))
#endif

#endif /* _NAN_H */

Will the above work in most cases? I tried log(0), 1.0/0.0, and 1.0e
+300. They seemed to work fine.

 

[Keith Thompson]

Collecting ideas from you, I wrote a header file to handle the isnan
and isinf problem.

#ifndef _NAN_H
#define _NAN_H

#ifndef isnan
#define isnan(x) ((x) != (x))
#endif

#ifndef isinf
#define isinf(x) (((x) == (x)) && ((x) != 0) && ((x)/2 == (x)))
#endif

#endif /* _NAN_H */

Will the above work in most cases? I tried log(0), 1.0/0.0, and 1.0e
+300. They seemed to work fine.

I don't know of any problems, but as Gordon Burditt pointed out, you
need to check for corner cases. However, I'd suggest picking
different names, since "isnan" and "isinf" are defined as macro names
in <math.h> in C99. Also, the identifier "_NAN_H" is reserved to the
implementation. In general, it's best to avoid identifiers that start
with underscores.

If you're going to be posting here, you should learn some of the
guidelines. Please don't top-post; your response goes after, or
interspersed with, the text you're responding to. See:
http://www.caliburn.nl/topposting.html
http://www.cpax.org.uk/prg/writings/topposting.php

Trim quoted material down to what's necessary for your response to
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And please don't delete attribution lines (you didn't do so this time,
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