Question

Jon Skeet reports today (source) that :

Math.Max(1f, float.NaN) == NaN
new[] { 1f, float.NaN }.Max() == 1f

Why?

Edit: same issue with double also!

Answer 1

As others have posted, I tweeted one sort of "why" – in that it’s using IComparable as documented.

That just leads to another "why" though. In particular:

Console.WriteLine(Math.Max(0, float.NaN));  // Prints NaN
Console.WriteLine(0f.CompareTo(float.NaN)); // Prints 1

The first line suggests that NaN is regarded as being greater than 0. The second line suggests that 0 is regarded as being greater than NaN. (Neither of these can report the result of "this comparison doesn’t make sense", of course.)

I have the advantage of seeing all the reply tweets, of course, including these two:

It may seem unusual, but that’s the right answer. max() of an array is NaN iff all elements are NaN. See IEEE 754r.

Also, Math.Max uses IEEE 754r total ordering predicate, which specifies relative ordering of NaN vs. others.