In Scala, why is NaN not being picked up by pattern matching?

问题

My method is as follows

  def myMethod(myDouble: Double): Double = myDouble match {
    case Double.NaN => ...
    case _ => ...
  }

The IntelliJ debugger is showing NaN but this is not being picked up in my pattern matching. Are there possible cases I am omitting

回答1:

It is a general rule how 64-bit floating point numbers are compared according to IEEE 754 (not Scala or even Java related, see NaN):

double n1 = Double.NaN;
double n2 = Double.NaN;
System.out.println(n1 == n2);     //false

The idea is that NaN is a marker value for unknown or indeterminate. Comparing two unknown values should always yields false as they are well... unknown.

---

If you want to use pattern matching with NaN, try this:

myDouble match {
    case x if x.isNaN => ...
    case _ => ...
}

But I think pattern matching will use strict double comparison so be careful with this construct.

回答2:

You can write an extractor (updated according to bse's comment):

object NaN {
  def unapply(d:Double) = d.isNaN
}


0.0/0.0 match {
  case NaN() => println("NaN")
  case x => println("boring " + x)
}
//--> NaN

回答3:

Tomasz is correct. You should use isNaN instead.

scala> Double.NaN.isNaN
res0: Boolean = true

来源：https://stackoverflow.com/questions/6908252/in-scala-why-is-nan-not-being-picked-up-by-pattern-matching