斯卡拉鸭打字模式匹配

Question

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)
    ...
}

Answer 1

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match在多个位置使用相同内容并希望避免代码重复,则可能会有所帮助.

以下自定义不适用:

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;)