I have this tiny piece of code
double s = -2.6114289999999998;
double s7 = Math.Round(s, 7);
double s5 = Math.Round(s, 5);
double s6 = Math.Round(s, 6);
According to the CSharp language specification 3.0 the following might be the case:
See chapter 4.1.6 at the specification for more information
Floating-point operations may be performed with higher precision than the result type of the operation. For example, some hardware architectures support an “extended” or “long double” floating-point type with greater range and precision than the double type, and implicitly perform all floating-point operations using this higher precision type. Only at excessive cost in performance can such hardware architectures be made to perform floating-point operations with less precision, and rather than require an implementation to forfeit both performance and precision, C# allows a higher precision type to be used for all floating-point operations. Other than delivering more precise results, this rarely has any measurable effects. However, in expressions of the form x * y / z, where the multiplication produces a result that is outside the double range, but the subsequent division brings the temporary result back into the double range, the fact that the expression is evaluated in a higher range format may cause a finite result to be produced instead of an infinity.
In short: The C# spec actually states that hardware architectures may have some impact on floating point types (Double, Float).
The language specification as .doc can be found here.