Math "pow" in Java and C# return slightly different results?

Question

I am porting program from C# to java. I've faced a fact that

Java

Math.pow(0.392156862745098,1./3.) = 0.7319587495200227

C#

Math.Pow( 0.392156862745098, 1.0 / 3.0) =0.73195874952002271

this last digit leads to sufficient differences in further calculations. Is there any way to emulate c#'s pow?

Thanx

Answer 1

Just to confirm what Chris Shain wrote, I get the same binary values:

// Java
public class Test
{
    public static void main(String[] args)
    {
        double input = 0.392156862745098;
        double pow = Math.pow(input, 1.0/3.0);            
        System.out.println(Double.doubleToLongBits(pow));
    }
}

// C#
using System;

public class Test
{
    static void Main()
    {
        double input = 0.392156862745098;
        double pow = Math.Pow(input, 1.0/3.0);            
        Console.WriteLine(BitConverter.DoubleToInt64Bits(pow));
    }
}

Output of both: 4604768117848454313

In other words, the double values are exactly the same bit pattern, and any differences you're seeing (assuming you'd get the same results) are due to formatting rather than a difference in value. By the way, the exact value of that double is

0.73195874952002271118800535987247712910175323486328125

Now it's worth noting that distinctly weird things can happen in floating point arithmetic, particularly when optimizations allow 80-bit arithmetic in some situations but not others, etc.

As Henk says, if a difference in the last bit or two causes you problems, then your design is broken.

Answer 2

If your calculations are sensitive to this kind of difference then you will need other measures (a redesign).