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V3076. Comparison with 'double.NaN' is …
Analyzer Diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Diagnosis of micro-optimizations (C++)
Diagnosis of 64-bit errors (Viva64, C++)
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Additional information
Contents

V3076. Comparison with 'double.NaN' is meaningless. Use 'double.IsNaN()' method instead.

March 4, 2016

The analyzer detected that a variable of type float or double is compared with a float.NaN or double.NaN value. As stated in the documentation, if two double.NaN values are tested for equality by using the == operator, the result is false. So, no matter what value of type double is compared with double.NaN, the result is always false.

Consider the following example:

void Func(double d) {
  if (d == double.NaN) {
    ....
  }
}

It's incorrect to test the value for NaN using operators == and !=. Instead, method float.IsNaN() or double.IsNaN() should be used. The fixed version of the code:

void Func(double d) {
  if (double.IsNaN(d)) {
    ....
  }
}

This diagnostic is classified as:

You can look at examples of errors detected by the V3076 diagnostic.

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