comparing double value

[lupien1]

All variable, in the following "for loop exemple" except nbval are double type.

I have a problem with the condition "val <= valmax". When val is increment by the step value and equal to valmax the for loop break has if the val was greater than valmax. How can i correct this problem?

for (nbval = 0, val = val0; val <= valmax ; val += step)
{
values[nbval] = val;
nbval++;
}

Thank you and please excuse my english!!!

 

[DaveSinkula]

Equality and floating point don't always mix well.

http://www.eskimo.com/~scs/c-faq/q14.5.html

Maybe try something like this.

for ( nbval = 0, val = val0; val [blue]<[/blue] valmax [blue]+ step / 2[/blue]; val += step )

 

[ArkM]

Try this function:

void Dbl()
{
  double x = 0.0;
  for (int i = 0; i < 10; ++i)
      x += 0.1;
  if (x == 1.0)
     printf("It's impossible!\n");
  else
     printf("10*0.1 == %20.16f (he-he)\n",x);
}

Never compare floating point number on equality. Doubles are approximations of numbers, not math real numbers.

 

[lupien1]

Thank you both for the time you took to answer me.

We add the same problem with our fortran code and we did the same thing (val < valmax + step / 1000) to correct it. I was hoping that a special configuration can solve this problem.

Thank's again

 

[cperl]

excellent idea.

 

[ArkM]

No any silver bullets (soft or hard to solve this problem. It's an old and well-known computational mathematics issue. All your math methods must recognize and take into account this problem (in Fortran, C, C++, asm etc...
Get a good book about comp math...

 

[lionelhill]

Usual way to deal with equality of any floating point thing is to test +/- an acceptable error. My favourite maths methods book is Sedgwick et al. Numerical Methods in C.

 

[OlafBogus]

The usual way is this

(fabs(Val1 - Val2) < 0.0001)

adjusting the number on the right for precision.