Why does Convert.ToDecimal(Double) round to 15 significant figures?

Question

I have a double with 17 digits after the decimal point, i.e.:

double myDouble = 0.12345678901234567;

If I convert this to a decimal like this:

decimal myDecimal = Convert.ToDecimal(myDouble);

then the value of myDecimal is rounded, as per the Convert.ToDecimal documentation, to 15 digits (i.e. 0.0123456789012345). My question is, why is this rounding performed?

I understand that if my original number could be accurately represented in base 10 and I was trying to store it as a double, then we could only have confidence in the first 15 digits. The final two digits would be subject to rounding error. But, that's a base 10 biased point of view. My number may be more accurately represented by a double and I wish to convert it to decimal while preserving as much accuracy as possible.

Shouldn't Convert.ToDecimal aim to minimise the difference between myDouble and (double)Convert.ToDecimal(myDouble)?

Answer 1

From the documentation of Double:

A Double value has up to 15 decimal digits of precision, although a maximum of 17 digits is maintained internally

So, as the double value itself has a maximum of 15 decimal places, converting it to Decimal will result in a Decimal value with 15 significant figures.