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Home/ Questions/Q 3605404
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T21:07:10+00:00 2026-05-18T21:07:10+00:00

I stumbled upon .NET’s definition of double.NaN in code: public const double NaN =

  • 0

I stumbled upon .NET’s definition of double.NaN in code:

public const double NaN = (double)0.0 / (double)0.0;

This is done similarly in PositiveInfinity and NegativeInfinity.

double.IsNaN (with removing a few #pragmas and comments) is defined as:

[Pure] 
[ReliabilityContract(Consistency.WillNotCorruptState, Cer.Success)]
public static bool IsNaN(double d) 
{
    if (d != d)
    { 
        return true; 
    }
    else 
    {
        return false;
    }
}

This is very counter-intuitive to me.

Why is NaN defined as division by zero? How is 0.0 / 0.0 represented “behind the scenes”? How can division by 0 be possible in double, and why does NaN != NaN?

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  1. Editorial Team
    Editorial Team
    2026-05-18T21:07:11+00:00Added an answer on May 18, 2026 at 9:07 pm

    Fairly simple answer here. .Net framework has implemented the floating point standard specified by the IEEE (System.Double complies with the IEC 60559:1989 (IEEE 754) standard for binary floating-point arithmetic.). This is because floating point arithmetic actually has to work across many systems, not just x86/64 architectures, so by following the conventions this ensures that there will be less compatibility issues (for instance porting code from a DSP into an x86 processor).

    As for the d != d, this is a performance optimisation. Basically this instruction relies on a hardware instruction which can very quickly determine if two double floating point numbers are equal. Under the standard, NAN != NAN and therefore is the fastest way to test. Trying to find a reference for you.

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