Question

我的addvf函数会根据added键将value_name参数添加到地图中的值存储区。

我期待看到打印输出“你还没有煮咖啡”和

的打印输出

println(values("sugar_in_coffee"))

是1。

addvf("sugar_in_coffee")只应在values("coffee")为＆gt; = 1时调用，如readvf函数所示。相反，这个addvf被称为声明它。

如果我使用() =>添加addvf，则acts("add_sugar_to_coffee")()不返回任何内容。

如何阻止评估作为参数传递的函数？

import scala.collection.mutable.HashMap

object dumbhash {

 val actions = new HashMap[String, ()=>Unit]

 def e(action_name: String) = {
   actions(action_name)()
 }

  val values = new HashMap[String, Int]
    values += "coffee" -> 0
    values += "bread" -> 0
    values += "sugar_in_coffee" -> 0
    println(values("coffee"))

def readf(value_name: String, sufficient:Int, t:()=>Unit, f:() => Unit): ()=>Unit = {
    if (values(value_name) >= sufficient) t else f
}

def addvf(value_name: String, added:Int, f:() =>Unit): ()=>Unit = {
   val v = values(value_name)
   values(value_name) = v + added
   f
}

val acts = new HashMap[String, ()=> Unit]

def desc(str: String) {
  println(str)
}


values("coffee") = 0
values("sugar_in_coffee") = 1
values += "made_coffee" -> 0

acts += "check_coffee" -> readf("coffee", 1,
        readf("sugar_in_coffee", 1, () => desc("The coffee smells brown, like you spilled too much sugar in the cup."),
        () => desc("Strong as love black coffee, waking you into this morning." )),

         readf("made_coffee", 1, () => desc("You drained the cup."), () => desc("You're the type who needs coffee to make coffee")))

acts("check_coffee")()

acts += "add_sugar_to_coffee" -> readf("made_coffee", 1, addvf("sugar_in_coffee", 8, () => desc("You dump sugar into the raspberry tree coffee mug.")),
 () => desc("You haven't made coffee yet"))

acts("add_sugar_to_coffee")()
println(values("sugar_in_coffee")) //returns 9!


}

Answer 1

尝试将addvf定义为

def addvf(value_name: String, added:Int, f:() =>Unit): ()=>Unit = () => {
   val v = values(value_name)
   values(value_name) = v + added
   f()
}

顺便说一句，你真正需要的是一个状态monad）请参阅http://eed3si9n.com以供参考）

如何防止作为参数传递的函数进行评估？

1 个答案: