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scala - Scala：泛型和隐式

问题描述

我必须开发类 StackMachine[T]。如果 T = Boolean，那么应该有逻辑运算。如果 T = Int,Double,Long 等，应该有算术运算。首先我开发了类 Stack[T]。

class Stack[T](val stack: List[T]) {
    val length: Int = stack.length
    def isEmpty: Boolean = {length == 0}
    def push(x: T): Stack[T] = {
      new Stack[T](x :: stack)
    }

     def peak: T = {
       if (this.isEmpty)
         throw new ArrayIndexOutOfBoundsException
         else stack.head
     }
    def pop(): Stack[T] = {
      if (this.isEmpty)
        throw new ArrayStoreException()
      val x :: xs = stack
      new Stack[T](xs)
    }

薄是我不知道如何开发 StackMachine[T] 取决于类型的操作的存在。我试过这个：

case class StackMachine[T](val stack:Stack[T]){
    def const(x: T): StackMachine[T] = {new StackMachine[T](new Stack[T](this.stack.push(x).stack))}
    def dup: StackMachine[T] = {new StackMachine[T](new Stack[T](this.stack.push(this.stack.peak).stack))}
    def swap: StackMachine[T] = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      val secondPeak = secondStack.peak
      val finalStack = secondStack.pop().push(startPeak)
      StackMachine[T](stack)
    }
    def and(): StackMachine[Boolean] = {
      val startStack = this.stack.asInstanceOf[Stack[Boolean]]
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      val secondPeak = secondStack.peak
      StackMachine[Boolean](new Stack[Boolean](secondStack.push(startPeak && secondPeak).stack))

    }

    def or: StackMachine[Boolean] = {
      val startStack = this.stack.asInstanceOf[Stack[Boolean]]
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      val secondPeak = secondStack.pop().peak
      StackMachine[Boolean](new Stack[Boolean](secondStack.push(startPeak || secondPeak).stack))
    }

    def xor: StackMachine[Boolean] = {
      val startStack = this.stack.asInstanceOf[Stack[Boolean]]
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      val secondPeak = secondStack.pop().peak
      StackMachine[Boolean](new Stack[Boolean](secondStack.push(startPeak ^ secondPeak).stack))
    }

    def sum(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.plus(startPeak,input)).stack))
    }

    def dif(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.minus(startPeak,input)).stack))
    }

    def mul(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.toDouble(startPeak).*(N.toDouble(input)).asInstanceOf[T]).stack))
    }

    def div(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.toDouble(startPeak)./(N.toDouble(input)).asInstanceOf[T]).stack))
    }

    def min(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.min(startPeak,input)).stack))
    }
    def max(input : T)(implicit N: Numeric[T])  = {
      val startStack = this.stack
      val startPeak = startStack.peak
      val secondStack = startStack.pop()
      StackMachine[T](new Stack[T](secondStack.push(N.max(startPeak,input)).stack))
    }



  }

但这是错误的，因为操作不应该有输入参数，因为所有变量都必须从堆栈中获取。不仅如此，这样我就无法创建 diff 和 mul 函数。我想使 StackMachine[T] 抽象并使用隐式对象，但失败了，因为在这种情况下，我的函数无法返回 StackMachine。可能是我对隐含的理解不够好，还是有另一种方法可以做到这一点？

标签： scalagenericsimplicit

解决方案

然后是的，似乎该项目打算使用typeclass来解决。

例如，请参见Boolean -like这个小的and：

sealed trait BehavesAsBoolean[T] {
  def and(t1: T, t2: T): T
}

object BehavesAsBoolean {
  implicit final val BooleanBehavesAsBoolean: BehavesAsBoolean[Boolean] =
    new BehavesAsBoolean[Boolean] {
      override def and(b1: Boolean, b2: Boolean): Boolean =
        b1 && b2
    }
}

final class StackMachine[T](stack: Stack[T]) {
  def and(implicit ev: BehavesAsBoolean[T]): Option[StackMachine[T]] =
    for {
      // I changed the implementation of pop to return an Option[(T, Stack[T])]
      (b1, s2) <- stack.pop
      (b2, s3) <- s2.pop
    } yield {
      new StackMachine(s3.push(ev.and(b1, b2)))
    }
}

当然，您可能仍然更喜欢抛出异常而不是使用Option
反正，我希望这可以帮助您完成代码。

您可以看到这里运行的代码。

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