我有一个类似下面的案例类:
// parent class
sealed abstract class Exp()
// the case classes I want to match have compatible constructors
case class A (a : Exp, b : Exp) extends Exp
case class B (a : Exp, b : Exp) extends Exp
case class C (a : Exp, b : Exp) extends Exp
// there are other case classes extending Exp that have incompatible constructor, e.g.
// case class D (a : Exp) extends Exp
// case class E () extends Exp
// I don't want to match them
我想要匹配:
var n : Exp = ...
n match {
...
case e @ A (a, b) =>
foo(e, a)
foo(e, b)
case e @ B (a, b) =>
foo(e, a)
foo(e, b)
case e @ C (a, b) =>
foo(e, a)
foo(e, b)
...
}
def foo(e : Exp, abc : Exp) { ... }
有没有办法将这三种情况合并为一个案例(不向A,B,C添加中间父类)?我无法更改A,B,C或Exp的定义。某种:
var n : Exp = ...
n match {
...
case e @ (A | B | C) (a, b) => // invalid syntax
foo(e, a)
foo(e, b)
...
}
这显然不起作用,也没有:
var n : Exp = ...
n match {
...
case e @ (A (a, b) | B (a, b) | C (a, b)) => // type error
foo(e, a)
foo(e, b)
...
}
答案 0 :(得分:11)
虽然以下“解决方案”实际上只是编写已有内容的另一种方式,但如果您需要在多个位置使用相同的match
并希望避免代码重复,则可能会有所帮助。< / p>
以下自定义取消应用:
object ExpABC {
def unapply(e:Exp):Option[(Int, Int)] = e match {
case A(a, b) => Some(a, b)
case B(a, b) => Some(a, b)
case C(a, b) => Some(a, b)
case _ => None
}
}
允许你写
n match {
case e @ ExpABC(a, b) =>
println(e)
println(a)
println(b)
}
这样您根本不需要修改原始类。我不知道更好的方法来做这个不涉及修改A / B / C类,但我渴望学习@ Stackoverflow;)