scala - 更新不可变对象(immutable对象)

Question

我构建了以下类:

class Player(val name: String, val onField: Boolean, val draft: Int, val perc: Int, val height: Int, val timePlayed: Int) {
override def toString: String = name

我正在努力

def play(team: List[Player]): List[Player] =
team map (p => new Player(p.name, p.onField, p.draft, p.perc, p.height, p.timePlayed + 1))

这实际上是将字段“timePlayed”递增 1，并返回新的玩家“列表”。

有没有更方便的方法呢？也许:

def play(team: List[Player]): List[Player] =
team map (p => p.timeIncremented())

我的问题是如何以更方便的方式实现 timeIncremented()？这样我就不必做:

new Player(p.name, p.onField, p.draft, p.perc, p.height, p.timePlayed + 1)

谢谢!

Answer 1

您可以将 Player 定义为 case class 并使用编译器生成的方法 copy:

case class Player(name: String, onField: Boolean, draft: Int, perc: Int, height: Int, timePlayed: Int) {
    override def toString: String = name
}

def play(team: List[Player]): List[Player] =
    team map (p => p.copy(timePlayed = p.timePlayed + 1))

此外，如您所见，构造函数参数默认为 val。

并且您可以在 Player 中定义 timeIncremented 并完全按照您的需要使用它:

case class Player(name: String, onField: Boolean, draft: Int, perc: Int, height: Int, timePlayed: Int) {
    override def toString: String = name
    def timeIncremented: Player = copy(timePlayed = this.timePlayed + 1)
}

def play(team: List[Player]): List[Player] =
    team map (_.timeIncremented)

对于更复杂的情况，您可以查看镜头:
http://akisaarinen.fi/blog/2012/12/07/boilerplate-free-functional-lenses-for-scala/
Cleaner way to update nested structures