.NET / C# arithmetic deviation (addition)

Question

Summation of these numbers give different results in .NET Core / C# and on other compilers.

3987908.698692091 + 92933945.11382028 + 208218.11919727124 + 61185833.06829034

.NET Core / C# : 158315904.99999997

Others: 158315905

Clearly the .NET Core / C# one is deviated.

Here is the code in C#:

double[] no = {
    3987908.698692091,
    92933945.11382028,
    208218.11919727124,
    61185833.06829034
};

Console.WriteLine("{0}", no.Sum());

Here is the code in C++

vector<double> no = {
    3987908.698692091,
    92933945.11382028,
    208218.11919727124,
    61185833.06829034
    };
    
cout << setprecision(16);
cout << "sum: " << sum(no) << " \n";

double sum(vector<double> &fa)
{
    double sum = 0.0;
    for(double f : fa)
        sum = sum + f;
    return sum;
}

PS: Using decimal also gives the same outcome. I believe mono C# compiler might give the same result as the C++ one.

Is there a way to fix the deviation problem either with compiler options or somehow inside C#?

Answer 1

This is only affecting the formatted output - the underlying numbers have not changed.

This was changed with .NET Core 3.0

If you change the output to:

Console.WriteLine("{0:G15}", no.Sum());

the output will be the same for both .NET 4.8 and .NET Core 3.0 and later.

The change to the formatting that you're seeing was made to implement IEEE 754-2008 correctly.

In particular, this requirement mandated the change:

The standard requires operations to convert between basic formats and external character sequence formats. Conversions to and from a decimal character format are required for all formats. Conversion to an external character sequence must be such that conversion back using round to nearest, ties to even will recover the original number.

Obviously the default round-trip conversion wouldn't work for .NET Framework because a value of 158315904.99999997 was converted to a string "158315905" which, when converted back to a binary floating point number, would differ from the original.