Re: Bug in conversion to ingegers???

0.01F cannot be exactly represented as a floating point number. It is
actually equal to something like .00999999999etc. So, when you multiply it
by 100, you get .99999999999etc

As far as I can tell, the multiplcation is done using doubles, so
assigning that double to a float has it rounded to the nearest value in the
float range, which in this case, is 1.

However, when you cast it directly to an int, the decimal portion is
truncated, before it can be rounded up.

To demostrate this, change the line in the second block to

i = (int)(float) (100.0 * 0.01f);

And you will get the results you expect.

Just another of the reasons why floating point is evil.
--
--
Truth,
James Curran
[erstwhile VC++ MVP]

Home: www.noveltheory.com Work: www.njtheater.com
Blog: www.honestillusion.com Day Job: www.partsearch.com

"Vladimir" <xozar@xxxxxx> wrote in message
news:etjejFyTFHA.2172@xxxxxxxxxxxxxxxxxxxxxxx
> Can anybody say me why...
>
> /*
> int i;
> float f;
>
> // Result is one.
> f = 100 * 0.01f;
> i = (int)f;
> MessageBox.Show(i.ToString());
>
> // Result is zero!!!
> i = (int)(100 * 0.01f);
> MessageBox.Show(i.ToString());
> */
>
>
>
>
>
>
>


.

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