c++ not how do I make a portable isnan/isinf function




how to avoid nan in c (9)

I've not tried this, but I would think

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

would work. It feels like there should be a better way for isinf, but that should work.

I've been using isinf, isnan functions on Linux platforms which worked perfectly. But this didn't work on OS-X, so I decided to use std::isinf std::isnan which works on both Linux and OS-X.

But the Intel compiler doesn't recognize it, and I guess its a bug in the intel compiler according to http://software.intel.com/en-us/forums/showthread.php?t=64188

So now I just want to avoid the hassle and define my own isinf, isnan implementation.

Does anyone know how this could be done?

edit:

I ended up doing this in my source code for making isinf/isnan working

#include <iostream>
#include <cmath>

#ifdef __INTEL_COMPILER
#include <mathimf.h>
#endif

int isnan_local(double x) { 
#ifdef __INTEL_COMPILER
  return isnan(x);
#else
  return std::isnan(x);
#endif
}

int isinf_local(double x) { 
#ifdef __INTEL_COMPILER
  return isinf(x);
#else
  return std::isinf(x);
#endif
}


int myChk(double a){
  std::cerr<<"val is: "<<a <<"\t";
  if(isnan_local(a))
    std::cerr<<"program says isnan";
  if(isinf_local(a))
    std::cerr<<"program says isinf";
  std::cerr<<"\n";
  return 0;
}

int main(){
  double a = 0;
  myChk(a);
  myChk(log(a));
  myChk(-log(a));
  myChk(0/log(a));
  myChk(log(a)/log(a));

  return 0;
}

Answer #1

This works under Visual Studio 2008:

#include <math.h>
#define isnan(x) _isnan(x)
#define isinf(x) (!_finite(x))
#define fpu_error(x) (isinf(x) || isnan(x))

For safety, I recommend using fpu_error(). I believe some numbers are picked up with isnan(), and some with isinf(), and you need both to be safe.

Here is some test code:

double zero=0;
double infinite=1/zero;
double proper_number=4;
printf("isinf(infinite)=%d.\n",isinf(infinite));
printf("isinf(proper_number)=%d.\n",isinf(proper_number));
printf("isnan(infinite)=%d.\n",isnan(infinite));
printf("isnan(proper_number)=%d.\n",isnan(proper_number));

double num=-4;
double neg_square_root=sqrt(num);
printf("isinf(neg_square_root)=%d.\n",isinf(neg_square_root));
printf("isinf(proper_number)=%d.\n",isinf(proper_number));
printf("isnan(neg_square_root)=%d.\n",isnan(neg_square_root));
printf("isnan(proper_number)=%d.\n",isnan(proper_number));

Here is the output:

isinf(infinite)=1.
isinf(proper_number)=0.
isnan(infinite)=0.
isnan(proper_number)=0.
isinf(neg_square_root)=1.
isinf(proper_number)=0.
isnan(neg_square_root)=1.
isnan(proper_number)=0.

Answer #2

As brubelsabs said Boost offers this feature but, as reported here, instead of using

if (boost::math::isnan(number))

This should be used:

if ((boost::math::isnan)(number))

Answer #3

Just use that super simple IEEE 754-1985-compliant code:

static inline bool  ISINFINITE( float a )           { return (((U32&) a) & 0x7FFFFFFFU) == 0x7F800000U; }
static inline bool  ISINFINITEPOSITIVE( float a )   { return (((U32&) a) & 0xFFFFFFFFU) == 0x7F800000U; }
static inline bool  ISINFINITENEGATIVE( float a )   { return (((U32&) a) & 0xFFFFFFFFU) == 0xFF800000U; }
static inline bool  ISNAN( float a )                { return !ISINFINITE( a ) && (((U32&) a) & 0x7F800000U) == 0x7F800000U; }
static inline bool  ISVALID( float a )              { return (((U32&) a) & 0x7F800000U) != 0x7F800000U; }

Answer #4

You could also use boost for this task:

#include <boost/math/special_functions/fpclassify.hpp> // isnan

if( boost::math::isnan( ... ) .... )

Answer #5

According to this, infinity is easy to check:

  • sign = either 0 or 1 bit indicating positive/negative infinity.
  • exponent = all 1 bits.
  • mantissa = all 0 bits.

NaN is a bit more complicated because it doesn't have a unique representation:

  • sign = either 0 or 1.
  • exponent = all 1 bits.
  • mantissa = anything except all 0 bits (since all 0 bits represents infinity).

Below is the code for double-precision floating-point case. Single-precision can be similarly written (recall that the exponent is 11-bits for doubles and 8-bits for singles):

int isinf(double x)
{
    union { uint64 u; double f; } ieee754;
    ieee754.f = x;
    return ( (unsigned)(ieee754.u >> 32) & 0x7fffffff ) == 0x7ff00000 &&
           ( (unsigned)ieee754.u == 0 );
}

int isnan(double x)
{
    union { uint64 u; double f; } ieee754;
    ieee754.f = x;
    return ( (unsigned)(ieee754.u >> 32) & 0x7fffffff ) +
           ( (unsigned)ieee754.u != 0 ) > 0x7ff00000;
}

The implementation is pretty straightforward (I took those from the OpenCV header files). It uses a union over an equal-sized unsigned 64-bit integer which you might need to correctly declare:

#if defined _MSC_VER
  typedef unsigned __int64 uint64;
#else
  typedef uint64_t uint64;
#endif

Answer #6

this works on osx

#include <math.h>

also this might be portable,

int isinf( double x ) { return x == x - 1; }

edit:

as Chris pointed out the above may fail with large x

int isinf( double x ) { return x == x * 2; }

Answer #7

No-one seems to have mentioned the C99 function fpclassify which returns:

One of FP_INFINITE, FP_NAN, FP_NORMAL, FP_SUBNORMAL, FP_ZERO or implementation-defined type, specifying the category of arg.

This works with visual studio, but I don't know about OS-X.


Answer #8

Well, ideally, you'd wait until Intel fixes the bug or provides a workaround :-)

But if you want to detect NaN and Inf from IEEE754 values, map it to an integer (32 or 64 bit depending on whether it's single or double precision) and check if the exponent bits are all 1. This indicates those two cases.

You can distinguish between NaN and Inf by checking the high order bit of the mantissa. If it's 1, that's NaN otherwise Inf.

+/-Inf is dictated by the sign bit.

For single precision (32-bit values), the sign is the high-order bit (b31), exponent is the next eight bits (plus a 23-bit mantissa). For double precision, the sign is still the high-order bit but the exponent is eleven bits (plus 52 bits for the mantissa).

Wikipedia has all the gory details.

The following code shows you how the encoding works.

#include <stdio.h>

static void decode (char *s, double x) {
    long y = *(((long*)(&x))+1);

    printf("%08x ",y);
    if ((y & 0x7ff80000L) == 0x7ff80000L) {
        printf ("NaN  (%s)\n", s);
        return;
    }
    if ((y & 0xfff10000L) == 0x7ff00000L) {
        printf ("+Inf (%s)\n", s);
        return;
    }
    if ((y & 0xfff10000L) == 0xfff00000L) {
        printf ("-Inf (%s)\n", s);
        return;
    }
    printf ("%e (%s)\n", x, s);
}

int main (int argc, char *argv[]) {
    double dvar;

    printf ("sizeof double = %d\n", sizeof(double));
    printf ("sizeof long   = %d\n", sizeof(long));

    dvar = 1.79e308; dvar = dvar * 10000;
    decode ("too big", dvar);

    dvar = -1.79e308; dvar = dvar * 10000;
    decode ("too big and negative", dvar);

    dvar = -1.0; dvar = sqrt(dvar);
    decode ("imaginary", dvar);

    dvar = -1.79e308;
    decode ("normal", dvar);

    return 0;
}

and it outputs:

sizeof double = 8
sizeof long   = 4
7ff00000 +Inf (too big)
fff00000 -Inf (too big and negative)
fff80000 NaN  (imaginary)
ffefdcf1 -1.790000e+308 (normal)

Just keep in mind that this code (but not the method) depends a great deal on the sizes of your longs which is not overly portable. But, if you have to bit-fiddle to get the information, you've already entered that territory :-)

As an aside, I've always found Harald Schmidt's IEEE754 converter very useful for floating point analysis.





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