Not a number problem

[istillshine@gmail.com]

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]

Re: Not a number problem

On 4 Apr, 15:43, istillsh...@gma il.com wrote:

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 (

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]

Re: Not a number problem

oops! last post sent in error

On 4 Apr, 15:43, istillsh...@gma il.com wrote:

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


--
Nick Keighley

[vippstar@gmail.com]

Re: Not a number problem

On Apr 4, 5:55 pm, Nick Keighley <nick_keighley_ nos...@hotmail. com>
wrote:

On 4 Apr, 15:43, istillsh...@gma il.com wrote:
>

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 (

Hmm.. something went wrong here? :-)
The macro is __STDC_IEC_559_ _
<snip>

[Philip Potter]

Re: Not a number problem

Nick Keighley wrote:

oops! last post sent in error
>
On 4 Apr, 15:43, istillsh...@gma il.com wrote:
>

>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.

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)

>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

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.

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

>
probably c90

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

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

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

>
yes

Probably.

[istillshine@gmail.com]

Re: Not a number problem

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 */
}
}



On Apr 4, 11:02 am, Nick Keighley <nick_keighley_ nos...@hotmail. com>
wrote:

oops! last post sent in error
>
On 4 Apr, 15:43, istillsh...@gma il.com wrote:
>

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
>
--
Nick Keighley

[jacob navia]

Re: Not a number problem

istillshine@gma il.com wrote:

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;
}


--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique

http://www.cs.virginia.edu/~lcc-win32

[Walter Roberson]

Re: Not a number problem

In article <ft693m$88u$2@a ioe.org>, jacob navia <jacob@nospam.o rgwrote:

>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.

--
"It's a hard life sometimes and the biggest temptation is to let
how hard it is be an excuse to weaken." -- Walter Dean Myers

[jacob navia]

Re: Not a number problem

Walter Roberson wrote:

In article <ft693m$88u$2@a ioe.org>, jacob navia <jacob@nospam.o rgwrote:
>

>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.
>

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.


--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique

http://www.cs.virginia.edu/~lcc-win32

[Keith Thompson]

Re: Not a number problem

roberson@ibd.nr c-cnrc.gc.ca (Walter Roberson) writes:

In article <ft693m$88u$2@a ioe.org>, jacob navia <jacob@nospam.o rgwrote:
>

>>P.S.
>>I think

>

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

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_n an(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.)

--
Keith Thompson (The_Other_Keit h) <kst-u@mib.org>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[istillshine@gmail.com]

Re: Not a number problem

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+0x3 6d):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;
}


On Apr 4, 6:12 pm, jacob navia <ja...@nospam.c omwrote:

istillsh...@gma il.com wrote:

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;
>
}
>
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatiquehtt p://www.cs.virginia .edu/~lcc-win32

[istillshine@gmail.com]

Re: Not a number problem

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.


On Apr 5, 12:18 am, istillsh...@gma il.com wrote:

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+0x3 6d):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;
>
}
>
On Apr 4, 6:12 pm, jacob navia <ja...@nospam.c omwrote:
>

istillsh...@gma il.com wrote:

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;

>

}

>

--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatiquehtt p://www.cs.virginia .edu/~lcc-win32

[Gordon Burditt]

Re: Not a number problem

>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);
>}

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.

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

Ok, but I think it comes up with lots of false positives.

Also, is there a reason you need that (x != 0) condition in there?
isinf(0.0) should fail the (x + 1 == x) condition.

[Richard Tobin]

Re: Not a number problem

In article <05adnaK7WogAh2 ranZ2dnUVZ_qiin Z2d@internetame rica>,
Gordon Burditt <gordonb.rmk8e@ burditt.orgwrot e:

> 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.

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

-- Richard
--
:wq

[istillshine@gmail.com]

Re: Not a number problem

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

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.