How to use xor to swap a float and an uint number?

Question

I am trying to swap the 2 numbers using xor(^). This is what i wrote so far:

            float X = 5.5f;
            uint Y = 10123456;
            X = X ^ Y;
            Y = Y ^ X;
            X = X ^ Y;

Answer 1

How to use xor to swap a float and an uint number?

You Can't Do That™.

XOR is a bitwise operator. Using it on a floating point number requires a serious deep understanding of the way IEEE-754 floating point is represented.

There was a time when programmers might have needed to know all this -- I once handled a requirement to convert IEEE float into DEC float and back -- but now there's hardware and fully-debugged runtime libraries for the purpose. Don't reinvent the flat tire.

Your task requires converting X to a uint and Y to a float as the values are swapped.

Answer 2

Doing so requires interpreting the 32 bit floating point number as a 32 bit integer. This typically involves unsafe code like so:

namespace XorSwap
{
    class Program
    {
        unsafe static void Main()
        {
            float X = 5.5f;
            uint* pX = (uint*) &X;
            uint Y = 10123456;
            *pX = *pX ^ Y;
            Y = Y ^ *pX;
            *pX = *pX ^ Y;
            Console.WriteLine($"X = {X}");
            Console.WriteLine($"Y = {Y}");
            Console.WriteLine(Math.Abs(X - 1.41859833465e-38) < float.Epsilon); // 10123456 is 1.41859833465e-38 in IEEE 754 hex representation
            Console.WriteLine(Y == 0x40b00000); // 5.5f is 0x40b00000 in IEEE 754 hex represenatation
            Console.ReadLine();
        }
    }
}

Note that this is hardly useful. Compilers will notice a typical 3-step-swap operation and generate efficient assembly code. Make your code readable instead of applying premature optimization.