ahmet alp balkan ahmet alp balkan - 1 month ago 13
C# Question

Why (int)double.NaN and (int)double.PositiveInfinity are 0?

in C#, if you

0/0
you get an exception.

But if you
0.0/0
or
0.0/0.0
you get
double.NaN
and
double.Infinity
, respectively.

but if you cast these results to
int
, you get 0.

> (int)double.PositiveInfinity
0
> (int)double.NaN
0


Why is this the case? Isn't the runtime supposed to give casting error because infinity is clearly not a zero?

Answer

It depends on what kind of context you're in. If you use a checked context, you'll get an exception. The relevant section of the spec is section 6.2.1:

For a conversion from float or double to an integral type, the processing depends on the overflow checking context (ยง7.6.12) in which the conversion takes place:

  • In a checked context, the conversion proceeds as follows:
    • If the value of the operand is NaN or infinite, a System.OverflowException is thrown.
    • Otherwise, the source operand is rounded towards zero to the nearest integral value. If this integral value is within the range of the destination type then this value is the result of the conversion.
    • Otherwise, a System.OverflowException is thrown.
  • In an unchecked context, the conversion always succeeds, and proceeds as follows.
    • If the value of the operand is NaN or infinite, the result of the conversion is an unspecified value of the destination type.
    • Otherwise, the source operand is rounded towards zero to the nearest integral value. If this integral value is within the range of the destination type then this value is the result of the conversion.
    • Otherwise, the result of the conversion is an unspecified value of the destination type.

So in an unchecked context, the answer isn't necessarily 0 - it's an unspecified int value. In fact, in my testing, it comes up as int.MinValue rather than 0 in an unchecked context.

But fundamentally, if you want to do checking, use a checked context (at least for that expression).