Comparison Against NaN

[better_cs_now@yahoo.com]

Hello all,

The code below outputs "FALSE" on my platform. Does the language
guarantee this behavior?

Thanks,
Dave


#include <iostream>
#include <limits>

using namespace std;

int main()
{
double Result;

try
{
Result = numeric_limits< double>::quiet_ NaN(); // 1.#QNAN

if (Result == numeric_limits< double>::quiet_ NaN())
cout << "TRUE" << endl;
else
cout << "FALSE" << endl;
}
catch(...)
{
cout << "Exception! !!" << endl;
}
}

[robertwessel2@yahoo.com]

Re: Comparison Against NaN

On Jul 10, 2:33 pm, "better_cs_...@ yahoo.com"
<better_cs_...@ yahoo.comwrote:

Hello all,
>
The code below outputs "FALSE" on my platform. Does the language
guarantee this behavior?
>
Thanks,
Dave
>
#include <iostream>
#include <limits>
>
using namespace std;
>
int main()
{
        double Result;
>
        try
        {
                Result = numeric_limits< double>::quiet_ NaN(); // 1.#QNAN
>
                if (Result == numeric_limits< double>::quiet_ NaN())
                        cout << "TRUE" << endl;
                else
                        cout << "FALSE" << endl;
        }
        catch(...)
        {
                cout << "Exception! !!" << endl;
        }
>
>
>
}

C99 does if the implementation defines __STDC_IEC_559_ _. C89/90 says
nothing about that, although your implementation might, nor does C99
if that macro is not defined.

[robertwessel2@yahoo.com]

Re: Comparison Against NaN

On Jul 10, 3:59 pm, "robertwess...@ yahoo.com"
<robertwess...@ yahoo.comwrote:

C99 does if the implementation defines __STDC_IEC_559_ _.  C89/90 says
nothing about that, although your implementation might, nor does C99
if that macro is not defined.

And for some reason I though I was replying in comp.lang.c, not .c+
+...

Anyway, the current C++ standard is in the same boat as C89/90,
although the next standard will (likely) include the C99 IEEE math
support option.

[joseph cook]

Re: Comparison Against NaN

On Jul 10, 3:33 pm, "better_cs_...@ yahoo.com"
<better_cs_...@ yahoo.comwrote:

Hello all,
>
The code below outputs "FALSE" on my platform. Does the language
guarantee this behavior?

<snip>

No,
But the general idea of all types of NaN is that they always
compare false, even to each other.

Joe Cook

[Paavo Helde]

Re: Comparison Against NaN

"better_cs_now@ yahoo.com" <better_cs_now@ yahoo.comkirjut as:

Hello all,
>
The code below outputs "FALSE" on my platform. Does the language
guarantee this behavior?

If you ask so, then no (the presence of a NaN value is not required by
the standard). However, maybe you wanted to ask another question: does
the language guarantee that comparing NaN with itself yields false; in
this case, the answer is yes, these comparison rules are part of the
definition of a NaN.

hth
Paavo

>
Thanks,
Dave
>
>
#include <iostream>
#include <limits>
>
using namespace std;
>
int main()
{
double Result;
>
try
{
Result = numeric_limits< double>::quiet_ NaN(); // 1.#QNAN
>
if (Result == numeric_limits< double>::quiet_ NaN())
cout << "TRUE" << endl;
else
cout << "FALSE" << endl;
}
catch(...)
{
cout << "Exception! !!" << endl;
}
}