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
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.
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
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