Comparison Against NaN

[better_cs_now]

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]

On Jul 10, 2:33*pm, "[e-mail]"
<[e-mail]:

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]

On Jul 10, 3:59*pm, "[e-mail]"
<[e-mail]:

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]

On Jul 10, 3:33*pm, "[e-mail]"
<[e-mail]:

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]

"be***********@yahoo.com" <be***********@yahoo.comkirjutas:

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;
}
}