8
我有一个案例类像下面这样:斯卡拉鸭打字模式匹配
// 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) { ... }
有合并的方式,3箱子成一个单一的情况下(无需向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)
...
}
这显然是行不通的,也不做:
var n : Exp = ...
n match {
...
case e @ (A (a, b) | B (a, b) | C (a, b)) => // type error
foo(e, a)
foo(e, b)
...
}