Rene Brun wrote:
>
> This is the expected result.
> Replace
> a = (Double_t)((b-c)/b)
> by
> a = (Double_t)(b-c)/b
Maybe you should explain...
(Double_t)((b-c)/b) means: convert the result of the calculation (b-c)/b to
double.
But since b and c are both integers, the result will be integer too. And since
b-c is less than b the outcome is zero. An integer zero of course will be
transformed to a double zero.
But when you covert b-c to double BEFORE the division, the result will be a
double. Now you have a double (b-c) mixed with an integer (b), in such a case
the result will always be of the more precise type.
Ingo Strauch
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