Why does Math.Floor(Double) return a value of type Double?
c#, floor, math
Solution
The range of `double` is much wider than the range of `int` or `long`. Consider this code:
double d = 100000000000000000000d;
long x = Math.Floor(d); // Invalid in reality
The integer is outside the range of `long` - so what would you expect to happen?
Typically you know that the value will actually be within the range of `int` or `long`, so you cast it:
double d = 1000.1234d;
int x = (int) Math.Floor(d);
but the onus for that cast is on the developer, not on `Math.Floor` itself. It would have been unnecessarily restrictive to make it just fail with an exception for all values outside the range of `long`.
Problem
I need to get the left hand side integer value from a decimal or double. For Ex: I need to get the value 4 from 4.6. I tried using Math.Floor function but it's returning a double value, for ex: It's returning 4.0 from 4.6. The MSDN documentation says that it returns an integer value. Am I missing something here? Or is there a different way to achieve what I'm looking for?