Double to Decimal without rounding after 15 digits

Question

When converting a "high" precision Double to a Decimal I lose precision with Convert.ToDecimal or casting to (Decimal) due to Rounding.

Example :

double d = -0.99999999999999956d;
decimal result = Convert.ToDecimal(d); // Result = -1
decimal result = (Decimal)(d); // Result = -1

The Decimal value returned by Convert.ToDecimal(double) contains a maximum of 15 significant digits. If the value parameter contains more than 15 significant digits, it is rounded using rounding to nearest.

So I in order to keep my precision, I have to convert my double to a String and then call Convert.ToDecimal(String):

decimal result = System.Convert.ToDecimal(d.ToString("G20")); // Result = -0.99999999999999956d

This method is working but I would like to avoid using a String variable in order to convert a Double to Decimal without rounding after 15 digits?

Do you know in advance the range for d? I can offer some lightweight solutions in pseudocode (I'm not familiar enough with C# to write them in C#) if you know that d is in a range like [-2..2] — Pascal Cuoq
I'm assuming this is a double coming from somewhere else that you can't change? If it's declared as a decimal from the get-go (decimal d = -0.99999999999999956m;) it maintains that precision. — Chris Sinclair
You're right Chris, I cannot change the double value because it's return by another method that I cannot change — Steven Muhr
The MSDN help for Convert.ToDecimal says "The Decimal value returned by this method contains a maximum of 15 significant digits. If the value parameter contains more than 15 significant digits, it is rounded using rounding to nearest." — Matthew Strawbridge

Pascal Cuoq Pascal Cuoq · Accepted Answer · 2014-07-05T13:58:24

One possible solution is to decompose d as the exact sum of n doubles, the last of which is small and contains all the trailing significant digits that you desire when converted to decimal, and the first (n-1) of which convert exactly to decimal.

For the source double d between -1.0 and 1.0:

    decimal t = 0M;
    bool b = d < 0;
    if (b) d = -d;
    if (d >= 0.5) { d -= 0.5; t = 0.5M; }
    if (d >= 0.25) { d -= 0.25; t += 0.25M; }
    if (d >= 0.125) { d -= 0.125; t += 0.125M; }
    if (d >= 0.0625) { d -= 0.0625; t += 0.0625M; }
    t += Convert.ToDecimal(d);
    if (b) t = -t;

Test it on ideone.com.

Note that the operations d -= are exact, even if C# computes the binary floating-point operations at a higher precision than double (which it allows itself to do).

This is cheaper than a conversion from double to string, and provides a few additional digits of accuracy in the result (four bits of accuracy for the above four if-then-elses).

Remark: if C# did not allow itself to do floating-point computations at a higher precision, a good trick would have been to use Dekker splitting to split d into two values d1 and d2 that would convert each exactly to decimal. Alas, Dekker splitting only works with a strict interpretation of IEEE 754 multiplication and addition.

Another idea is to use C#'s version of frexp to obtain the significand s and exponent e of d, and to compute (Decimal)((long) (s * 4503599627370496.0d)) * <however one computes 2^e in Decimal>.

Double to Decimal without rounding after 15 digits

2 Answers