As you may know 0.1 = 1 X (10 ^-1) or 1e-1 and 0.01 = 1e-2, etc. So
1.082e-15 will be something like 0.000000000000001082 which is approx =
0. I think you are getting this result because of the precision of
double. Try using float.
What happens when you use C printf statements like:
printf( "x = %f", x );
About the loop stop early, doesnt make sense to me. You loop logic seems
to be ok. May be the output is still in buffer. Try printing somethin to
ouput before exiting.
Regards,
Saru
Rogelio Nodal wrote:
> Hello List:
>
> Is this a bug?Or is it me?
>
> [code]
>
> /* Created by Anjuta version 1.2.4 */
> /* This file will not be overwritten */
>
> #include <iostream>
> #include <fstream>
>
> using namespace std;
>
> int main()
> {
>
>
>
> double x = 0;
>
>
> x = -2.3;
> while(x <= 2.1)
> {
> cout << x << endl;
> x += 0.1;
> }
>
> return (0);
> }
>
> [code]
>
>
> When I run this little program I get the following output:
>
>
> -2.3
> -2.2
> -2.1
> -2
> -1.9
> -1.8
> -1.7
> -1.6
> -1.5
> -1.4
> -1.3
> -1.2
> -1.1
> -1
> -0.9
> -0.8
> -0.7
> -0.6
> -0.5
> -0.4
> -0.3
> -0.2
> -0.1
> 1.082e-15 ===> What the hell?
> 0.1
> 0.2
> 0.3
> 0.4
> 0.5
> 0.6
> 0.7
> 0.8
> 0.9
> 1
> 1.1
> 1.2
> 1.3
> 1.4
> 1.5
> 1.6
> 1.7
> 1.8
> 1.9
> 2 === Why did it stop here?It is supposed to stop at 2.1.
>
>
>
> Thanks a lot for your time.
>
>
> Rogelio.
>
>