如何使Scala重复控制直到抽象？ [英] How Does One Make Scala Control Abstraction in Repeat Until?

问题描述

我是Peter Pilgrim。我看到Martin Odersky在Scala中创建了控件抽象。但是，我似乎还不能在IntelliJ IDEA 9中重复它。是IDE吗？

I am Peter Pilgrim. I watched Martin Odersky create a control abstraction in Scala. However I can not yet seem to repeat it inside IntelliJ IDEA 9. Is it the IDE?

package demo

class Control {

  def repeatLoop ( body: => Unit ) = new Until( body )

  class Until( body: => Unit ) {
    def until( cond: => Boolean ) {
      body;
      val value: Boolean = cond;
      println("value="+value)
      if ( value ) repeatLoop(body).until(cond)
      // if  (cond) until(cond)
    }
  }

  def doTest2(): Unit = {
    var y: Int = 1
    println("testing ... repeatUntil() control structure")
    repeatLoop {
      println("found y="+y)
      y = y + 1
    }
    { until ( y < 10 ) }
  }

}

错误消息为：

信息：编译完成，出现1条错误和0条警告

信息：1条错误

信息：0条警告

C：\Users\Peter\IdeaProjects\HelloWord\src\demo\Control.scala

错误：Error：第（57）行错误：Control.this.repeatLoop （{

scala.this.Predef.println（ found y =。+（y））;

y = y。+（1）

}）类型的Control.this.Until不会接受参数

repeatLoop {

Information:Compilation completed with 1 error and 0 warnings
Information:1 error
Information:0 warnings
C:\Users\Peter\IdeaProjects\HelloWord\src\demo\Control.scala
Error:Error:line (57)error: Control.this.repeatLoop({
scala.this.Predef.println("found y=".+(y));
y = y.+(1)
}) of type Control.this.Until does not take parameters
repeatLoop {

可以认为rried函数的主体可以返回一个表达式（y + 1的值），但是repeatUntil的声明主体参数明确指出可以忽略还是不忽略？

In the curried function the body can be thought to return an expression (the value of y+1) however the declaration body parameter of repeatUntil clearly says this can be ignored or not?

错误是什么意思？

推荐答案

这是没有 StackOverflowError 。

scala>   class ConditionIsTrueException extends RuntimeException
defined class ConditionIsTrueException

scala>   def repeat(body: => Unit) = new {
 |     def until(condition: => Boolean) = { 
 |       try {
 |         while(true) {
 |           body
 |           if (condition) throw new ConditionIsTrueException
 |         }   
 |       } catch {
 |         case e: ConditionIsTrueException =>
 |       }   
 |     
 |     }   
 |   }
repeat: (body: => Unit)java.lang.Object{def until(condition: => Boolean): Unit}

scala> var i = 0              
i: Int = 0

scala> repeat { println(i); i += 1 } until(i == 3)
0
1
2

scala> repeat { i += 1 } until(i == 100000)       

scala> repeat { i += 1 } until(i == 1000000)

scala> repeat { i += 1 } until(i == 10000000)

scala> repeat { i += 1 } until(i == 100000000)

scala> 

根据Jesper和Rex Kerr，这是一个没有例外的解决方案。

According to Jesper and Rex Kerr here is a solution without the Exception.

def repeat(body: => Unit) = new {
  def until(condition: => Boolean) = { 
    do {
      body
    } while (!condition)
  }   
}

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