在 Scala 中，如何重新分配元组值？
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In Scala, how can I reassign tuple values?
提问by deltanovember
I'm trying to do something like the following
我正在尝试执行以下操作
var tuple = (1, "test")
tuple._2 = "new"
However this does not compile it complains about val
然而，这不会编译它抱怨 val
回答by Rex Kerr
You can't reassign tuple values. They're intentionally immutable: once you have created a tuple, you can be confident that it will never change. This is very useful for writing correct code!
您不能重新分配元组值。它们是故意不可变的：一旦你创建了一个元组，你就可以确信它永远不会改变。这对于编写正确的代码非常有用！
But what if you want a different tuple? That's where the copy method comes in:
但是如果你想要一个不同的元组怎么办？这就是复制方法的用武之地：
val tuple = (1, "test")
val another = tuple.copy(_2 = "new")
or if you really want to use a var to contain the tuple:
或者如果您真的想使用 var 来包含元组：
var tuple = (1, "test")
tuple = tuple.copy(_2 = "new")
Alternatively, if you really, really want your values to change individually, you can use a case class instead (probably with an implicit conversion so you can get a tuple when you need it):
或者，如果您真的，真的希望您的值单独更改，则可以改用 case 类（可能使用隐式转换，以便您可以在需要时获得元组）：
case class Doublet[A,B](var _1: A, var _2: B) {}
implicit def doublet_to_tuple[A,B](db: Doublet[A,B]) = (db._1, db._2)
val doublet = Doublet(1, "test")
doublet._2 = "new"
回答by Colliot
You can wrapper the component(s) you need to modify in a case class with a varmember, like:
您可以使用var成员在案例类中包装您需要修改的组件，例如：
case class Ref[A](var value: A)
var tuple = (Ref(1), "test")
tuple._1.value = 2
println(tuple._1.value) // -> 2
