在 Scala 中，为什么模式匹配没有选择 NaN？
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In Scala, why is NaN not being picked up by pattern matching?
提问by deltanovember
My method is as follows
我的方法如下
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
IntelliJ 调试器显示 NaN，但这在我的模式匹配中没有得到。是否有我省略的可能情况
回答by Tomasz Nurkiewicz
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):
根据 IEEE 754 比较 64 位浮点数是一般规则（与 Scala 甚至 Java 无关，请参阅NaN）：
double n1 = Double.NaN;
double n2 = Double.NaN;
System.out.println(n1 == n2); //false
The idea is that NaNis a marker value for unknownor indeterminate. Comparing two unknown values should always yields falseas they are well... unknown.
这个想法是未知或不确定NaN的标记值。比较两个未知值应该总是产生收益，因为它们很好......未知。false
If you want to use pattern matching with NaN, try this:
如果你想使用模式匹配 with NaN，试试这个：
myDouble match {
case x if x.isNaN => ...
case _ => ...
}
But I think pattern matching will use strict double comparison so be careful with this construct.
但我认为模式匹配将使用严格的双重比较，所以要小心这个结构。
回答by Landei
You can write an extractor (updated according to bse's comment):
您可以编写一个提取器（根据 bse 的评论更新）：
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by hammar
Tomasz is correct. You should use isNaNinstead.
托马兹是对的。你应该isNaN改用。
scala> Double.NaN.isNaN
res0: Boolean = true
