C# rounding differently depending on platform?

Question

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);
         


        

Answer 1

On x64, the SSE2 FPU is used and on x86 the x87 FPU is used.

It is possible (although not advised) to change the x87 precision to be the same as the SSE2 precision (i.e., use lower precision).

You can do that via the _controlfp() API.

The following program demonstrates. Run this program in x86 mode and you'll see how the use of _controlfp(0x00030000, 0x00020000) causes the output to change to be similar to (but not quite the same!) as the x64 version:

using System;
using System.Runtime.InteropServices;

namespace ConsoleApp3
{
    class Program
    {
        static void Main()
        {
            double s = -2.6114289999999998;

            Console.WriteLine(Math.Round(s, 7).ToString("R")); // -2.611429

            if (!Environment.Is64BitProcess)
                _controlfp(0x00030000, 0x00020000);

            Console.WriteLine(Math.Round(s, 7).ToString("R")); // -2.61142897605896
        }

        [DllImport("msvcrt.dll", CallingConvention = CallingConvention.Cdecl)]
        static extern uint _controlfp(uint newcw, uint mask);
    }
}

However, you should not mess around with the FPU in this way (and if you do, you should revert to the previous setting asap).