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Scala编译器无法推断混合类型以进行模式匹配

发布时间:2020-12-16 19:14:06 所属栏目:安全 来源:网络整理
导读:我有一个关于有限排列集的代数群的用例.因为我想将这个组用于各种不相关的排列类,我想把它作为一个混合特性来做.这是我尝试的摘录 trait Permutation[P : Permutation[P]] { this: P = def +(that: P): P //final override def equals(that: Any) = ... //fi
我有一个关于有限排列集的代数群的用例.因为我想将这个组用于各种不相关的排列类,我想把它作为一个混合特性来做.这是我尝试的摘录

trait Permutation[P <: Permutation[P]] { this: P =>
  def +(that: P): P

  //final override def equals(that: Any) = ...
  //final override lazy val hashCode = ...

  // Lots of other stuff
}

object Permutation {
  trait Sum[P <: Permutation[P]] extends Permutation[P] { this: P =>
    val perm1,perm2: P

    // Lots of other stuff
  }

  private object Sum {
    def unapply[P <: Permutation[P]](s: Sum[P]): Some[(P,P)] = Some(s.perm1,s.perm2)
    //def unapply(s: Sum[_ <: Permutation[_]]): Some[(Permutation[_],Permutation[_])] = Some(s.perm1,s.perm2)
  }

  private def simplify[P <: Permutation[P]](p: P): P = {
    p match {
      case Sum(a,Sum(b,c)) => simplify(simplify(a + b) + c)

      // Lots of other rules

      case _ => p
    }
  }
}

在某个时间点,我想调用简化方法,以便使用代数公理简化组操作的表达式.使用模式匹配似乎是有意义的,因为有很多公理需要评估,语法简洁.但是,如果我编译代码,我得到:

error: inferred type arguments [P] do not conform to method unapply's type parameter bounds [P <: Permutation[P]]

我不明白为什么编译器无法正确推断类型,我不知道如何帮助它.实际上,在这种情况下,当模式匹配时,P的参数类型是无关紧要的.如果p是任何排列和,则模式应该匹配.返回类型仍然是P,因为转换仅通过在P上调用运算符来完成.

因此,在第二次尝试中,我交换了注释掉的unapply版本.但是,我从编译器(2.8.2)得到一个断言错误:

assertion failed: Sum((a @ _),(b @ _)) ==> Permutation.Sum.unapply(<unapply-selector>) <unapply> ((a @ _),(b @ _)),pt = Permutation[?>: Nothing <: Any]

有什么线索我怎么能让编译器接受这个?

提前致谢!

解决方法

经过两天的繁殖后,我终于找到了一个没有警告的编译解决方案并通过了我的规范测试.以下是我的代码的可编辑摘录,以显示所需内容.但是请注意,代码是无操作的,因为我遗漏了实际执行排列的部分:

/**
 * A generic mix-in for permutations.
 * <p>
 * The <code>+</code> operator (and the apply function) is defined as the
 * concatenation of this permutation and another permutation.
 * This operator is called the group operator because it forms an algebraic
 * group on the set of all moves.
 * Note that this group is not abelian,that is the group operator is not
 * commutative.
 * <p>
 * The <code>*</code> operator is the concatenation of a move with itself for
 * <code>n</code> times,where <code>n</code> is an integer.
 * This operator is called the scalar operator because the following subset(!)
 * of the axioms for an algebraic module apply to it:
 * <ul>
 * <li>the operation is associative,*     that is (a*x)*y = a*(x*y)
 *     for any move a and any integers x and y.
 * <li>the operation is a group homomorphism from integers to moves,*     that is a*(x+y) = a*x + a*y
 *     for any move a and any integers x and y.
 * <li>the operation has one as its neutral element,*     that is a*1 = m for any move a.
 * </ul>
 * 
 * @param <P> The target type which represents the permutation resulting from
 *        mixing in this trait.
 * @see Move3Spec for details of the specification.
 */
trait Permutation[P <: Permutation[P]] { this: P =>
  def identity: P

  def *(that: Int): P
  def +(that: P): P
  def unary_- : P

  final def -(that: P) = this + -that
  final def unary_+ = this

  def simplify = this

  /** Succeeds iff `that` is another permutation with an equivalent sequence. */
  /*final*/ override def equals(that: Any): Boolean // = code omitted
  /** Is consistent with equals. */
  /*final*/ override def hashCode: Int // = code omitted

  // Lots of other stuff: The term string,the permutation sequence,the order etc.
}

object Permutation {
  trait Identity[P <: Permutation[P]] extends Permutation[P] { this: P =>
    final override def identity = this

    // Lots of other stuff.
  }

  trait Product[P <: Permutation[P]] extends Permutation[P] { this: P =>
    val perm: P
    val scalar: Int

    final override lazy val simplify = simplifyTop(perm.simplify * scalar)

    // Lots of other stuff.
  }

  trait Sum[P <: Permutation[P]] extends Permutation[P] { this: P =>
    val perm1,perm2: P

    final override lazy val simplify = simplifyTop(perm1.simplify + perm2.simplify)

    // Lots of other stuff.
  }

  trait Inverse[P <: Permutation[P]] extends Permutation[P] { this: P =>
    val perm: P

    final override lazy val simplify = simplifyTop(-perm.simplify)

    // Lots of other stuff.
  }

  private def simplifyTop[P <: Permutation[P]](p: P): P = {
    // This is the prelude required to make the extraction work.
    type Pr = Product[_ <: P]
    type Su = Sum[_ <: P]
    type In = Inverse[_ <: P]
    object Pr { def unapply(p: Pr) = Some(p.perm,p.scalar) }
    object Su { def unapply(s: Su) = Some(s.perm1,s.perm2) }
    object In { def unapply(i: In) = Some(i.perm) }
    import Permutation.{simplifyTop => s}

    // Finally,here comes the pattern matching and the transformation of the
    // composed permutation term.
    // See how expressive and concise the code is - this is where Scala really
    // shines!
    p match {
      case Pr(Pr(a,x),y) => s(a*(x*y))
      case Su(Pr(a,Pr(b,y)) if a == b => s(a*(x + y))
      case Su(a,Su(b,c)) => s(s(a + b) + c)
      case In(Pr(a,x)) => s(s(-a)*x)
      case In(a) if a == a.identity => a.identity
      // Lots of other rules

      case _ => p
    }
  } ensuring (_ == p)

  // Lots of other stuff
}

/** Here's a simple application of the mix-in. */
class Foo extends Permutation[Foo] {
  import Foo._

  def identity: Foo = Identity
  def *(that: Int): Foo = new Product(this,that)
  def +(that: Foo): Foo = new Sum(this,that)
  lazy val unary_- : Foo = new Inverse(this)

  // Lots of other stuff
}

object Foo {
  private object Identity
  extends Foo with Permutation.Identity[Foo]

  private class Product(val perm: Foo,val scalar: Int)
  extends Foo with Permutation.Product[Foo]

  private class Sum(val perm1: Foo,val perm2: Foo)
  extends Foo with Permutation.Sum[Foo]

  private class Inverse(val perm: Foo)
  extends Foo with Permutation.Inverse[Foo]

  // Lots of other stuff
}

如您所见,解决方案是定义simplifyTop方法本地的类型和提取器对象.

我还提供了一个如何将这种混合应用于Foo类的小例子.正如您所看到的,Foo只不过是一个工厂,可以根据自己的类型进行组合排列.如果您有许多这样的课程,那将是一个很大的好处,否则这些课程是无关的.

<咆哮>

但是,我无法抗拒说Scala的类型系统非常复杂!我是一名经验丰富的Java库开发人员,对Java Generics非常熟练.然而,花了两天的时间才弄清楚六行代码和三种类型和对象定义!如果这不是出于教育目的,我会抛弃这种方法.

现在,我很想知道,由于这种复杂性,Scala不会成为编程语言方面的下一个重大事件.如果你是一个Java开发人员,现在对Java泛型感到有些不舒服(不是我),那么你会讨厌Scala的类型系统,因为它至少可以说是对Java泛型概念添加不变量,协变量和逆变量.

总而言之,Scala的类型系统似乎解决了比开发人员更多的科学家.从科学的角度来看,很好地推断一个程序的类型安全性.从开发人员的角度来看,弄清楚这些细节的时间是浪费,因为它使他们远离程序的功能方面.

没关系,我会继续使用Scala.模式匹配,混合和高阶函数的组合太强大,不容错过.但是,如果没有过于复杂的类型系统,我觉得Scala会更高效.

< /咆哮>

(编辑:李大同)

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