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

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?

From stackoverflow

• 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.

• 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.

  Jon Skeet : The OP is looking for a tiny but positive number. double.MinValue is a massive and negative number.

  Yossarian : Uh? I thought, tnat numeric_limits returns minimal value that can be represented, not minimal nonzero number..

  Jon Skeet : Take a look at the values given in the question: "std::numeric_limits::min() == 2.2250738585072014e-308"

• 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.

  sthiers : Exactly the kind of answer I was looking for. Thanks Jon