double.Epsilon vs. std::numeric_limits<double>::min()

Question

Why double.Epsilon != std::numeric_limits<double>::min()?

On my PC: double.Epsilon == 4.9406564584124654E-324 and is defined in .NET std::numeric_limits<double>::min() == 2.2250738585072014e-308

Is there a way to get 2.2250738585072014e-308 from .NET?

Answer 1

Well you could use C++/CLI to return the value:

double epsilon() { return std::numeric_limits<double>::min(); }

Why would you want to though? Why do they have to be the same? You should try to avoid skating on the edges of your floating point numbers.

Answer 2

Epsilon is the minimal possible difference between two doubles. (Edit: not exact, it's the minimal positive nonzero number in this case).

double.MinValue

is what you need.

Answer 3

They're different because double.Epsilon returns the smallest representable value. numeric_limits<double>::min() returns the smallest normalized value.

Basically double.Epsilon is the equivalent to numeric_limits<double>::denorm_min().

The easiest way of getting the equivalent in .NET is probably to work out the bit pattern for the minimal normalized number and use BitConverter.Int64BitsToDouble.