Re: Comparison Against NaN

From:

Paavo Helde <nobody@ebi.ee>

Newsgroups:

comp.lang.c++

Date:

Sun, 13 Jul 2008 12:02:03 -0500

Message-ID:

<Xns9ADACBCA91D5nobodyebiee@216.196.97.131>

"better_cs_now@yahoo.com" <better_cs_now@yahoo.com> kirjutas:

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;
     }
}