strtol with floating point

**Laserve** · 07-26-2004

I have a problem that I don't understand. consider the following code:

double Scale;
char *fmterror;

Scale = strtol(string, &fmterror, 10);

if ( *fmterror != '\0' || Scale == 0.0 )
Scale = 1.0;
else
Scale /= 100;

it was compiled with Builder5 on windows2000 and it gives a "floating point overflow" on the line "Scale /= 100". The debugger shows that Scale = 150 at that point and 150 / 100 seems all right to me.

When I saw this I noticed that the code does do strtol instead of strtod and after changing double Scale into float Scale it worked. double is 8 bytes on this compiler and float 4 bytes, long also 4 bytes.

What could be the problem with this code when Scale is double?

**quzah** · 07-26-2004

Well, since you're actually trying to use floating point numbers, how about dividing by one?

Code:

Scale /= 100.0;

Quzah.

**Laserve** · 07-26-2004

Yes, that also didnt fix the problem

**Hammer** · 07-26-2004

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
  double  d;
  char    *p;

  d = strtod("100.12345", &p);

  if (*p != '\0' || d == 0.0)
  {
    d = 1.0;
  }
  else
  {
    d /= 100.0;
  }
  
  printf ("%f\n", d);

  return(0);
}

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