请在下面找一个令我困惑的简短例子。
我必须承认,在Scala中操纵存在类型有一些困难。
如何解决类型不匹配线56? 提交者是OK类型_ $ 1,而提议者的类型是_ $ 1&lt ;: Individual
提前致谢,
class Individual(n: String) {
protected val name=n
var preferred: Individual = this
override def toString(): String=name
}
class Man(n: String) extends Individual(n) { }
class Woman(n: String) extends Individual(n) { }
class Marriage(m: Man, w: Woman){
private val man=m
private val woman=w
def this(w: Woman, m: Man) = this(m,w)
override def toString(): String = man+"--"+woman
}
class Matching(){
private var list: List[Marriage] = Nil
def add(m: Marriage): Unit = { list = m ::list }
override def toString(): String= {
var s: String = ""
for (elm<-list) s=s+elm+" "
return s
}
}
object Test{
protected var male = true
def main(args: Array[String]): Unit = {
val al = new Man("Al")
val bob = new Man("Bob")
val alice = new Woman("Alice")
val barbara = new Woman("Barbara")
al.preferred = alice
bob.preferred = barbara
alice.preferred = bob
barbara.preferred = al
val men = Set(al, bob)
val women = Set(alice, barbara)
val m = new Matching()
//var proposers=women
var proposers: Set[_ <:Individual] = Set[Individual]()
if (male) proposers = men
else proposers = women
while (!proposers.isEmpty) {
for(proposer <- proposers) {
val proposer=proposers.head
if (proposer.isInstanceOf[Man])
m.add(new Marriage(
proposer.asInstanceOf[Man],
proposer.preferred.asInstanceOf[Woman]
))
else
m.add(new Marriage(
proposer.asInstanceOf[Woman],
proposer.preferred.asInstanceOf[Man]
))
proposers-=proposer//There is an error here
}
}
println(m)
}
}
答案 0 :(得分:7)
这段代码很乱。它的格式很差,它混合了标签和空格,即使在功能性解决方案几乎不需要考虑的最简单的地方,它也会使用可变性。
此代码也不会在国际范围内扩展到可能存在同性婚姻的国家。
自上而下工作......
我怀疑你从不希望直接实例Individual
,只有Man
或Woman
。因此,代数数据类型更有意义,这是通过sealed trait
和case class
子类型完成的。
我还会删除preferred
属性,因为它可能导致循环引用。在不可变数据中处理这个问题超出了我愿意回答的水平。
sealed trait Individual {
def name: String
override def toString(): String=name
}
//as it's a case class, `name` becomes a val,
//which implements the abstract `def name` from the trait
case class Man(name: String) extends Individual
case class Woman(name: String) extends Individual
Marriage
也可以是一个案例类,让我们把笨拙的类参数重复放到val中 - 它只是毫无意义的样板。这也是将辅助构造函数移动到伴随对象中的工厂方法的好时机:
case class Marriage(man: Man, woman: Woman) {
override def toString(): String = man + "--" + woman
}
object Marriage {
def apply(w: Woman, m: Man) = new Marriage(m,w)
}
Matching
几乎没有意义,整个课程只是为了包裹List
?这种事情在前Generics Java中是有意义的,但现在不再有了。无论如何我会保留它(现在)所以我可以修复toString
实现,这是一个非常可变的并且没有充分理由使用return
:
case class Matching(){
private var list: List[Marriage] = Nil
def add(m: Marriage): Unit = { list ::= m }
override def toString() = list.mkString(" ")
}
最后,问题的“肉”。评论是内联的,但您会注意到我不需要(或使用)Matching
。它被最终的println
全部替换
object Test{
//better name, and a val (because it never changes)
protected val menPropose = true
def main(args: Array[String]): Unit = {
// `new` not required for case classes
val al = Man("Al")
val bob = Man("Bob")
val alice = Woman("Alice")
val barbara = Woman("Barbara")
// remember how preference was removed from `Individual`?
val mprefs = Map( al -> alice, bob -> barbara )
val fprefs = Map( alice -> bob, barbara -> al )
val men = Set(al, bob)
val women = Set(alice, barbara)
// nicely immutable, and using the returned value from if/else
val proposers = if (menPropose) men else women
// no while loop, name shadowing, or mutability.
// just a simple for-comprehension
val marriages = for(proposer <- proposers) yield {
//pattern-matching beats `isInstanceOf`... every time
proposer match {
case m: Man => Marriage(m, mprefs(m))
case f: Woman => Marriage(f, fprefs(f))
}
}
println(marriages mkString " ")
}
}
还有更多可以在这里完成的事情。什么是同性关系?如果两个或更多人共享相同的偏好怎么办?如果有人没有偏好怎么办?
我还可以将某人的偏好类型编码为Individual
个实例。但那会更先进。