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我正在尝试存在类型。

我正在玩一个函数，该函数需要一个序列，其中该序列的元素都是相同的类型。我有..

def bar[X](as: Seq[A[X]]) = true

在哪里 ...

// parametised type to use in the question
trait A[T]

然后我遇到了“forSome”语法，发现我可以用它实现相同的约束。

为了比较，我写了以下内容......

// useful types 
trait A[T]
class AI extends A[Int]
class AS extends A[String]

// define two functions that both have the same constraint.
// ie the arg must be a Sequence with all elements of the same parameterised type

def foo(as: Seq[A[X]] forSome { type X }) = true

def bar[X](as: Seq[A[X]]) = true

// these compile because all the elements are the same type (AI)
foo(Seq(new AI, new AI))
bar(Seq(new AI, new AI))

// both these fail compilation as expected because 
// the X param of X[A] is different (AS vs AI)
foo(Seq(new AI, new AS))
bar(Seq(new AI, new AS))

我想了解的是 - 我错过了什么吗？一个签名比另一个签名有什么好处。

一个明显的区别是编译错误是不同的。

scala> foo(Seq(new AI, new AS))
<console>:12: error: type mismatch;
 found   : Seq[A[_ >: String with Int]]
 required: Seq[A[X]] forSome { type X }

              foo(Seq(new AI, new AS))
                     ^

scala> bar(Seq(new AI, new AS))
<console>:12: error: no type parameters for method bar: (as: Seq[A[X]])Boolean e
xist so that it can be applied to arguments (Seq[A[_ >: String with Int]])
 --- because ---
argument expression's type is not compatible with formal parameter type;
 found   : Seq[A[_ >: String with Int]]
 required: Seq[A[?X]]
              bar(Seq(new AI, new AS))
              ^
<console>:12: error: type mismatch;
 found   : Seq[A[_ >: String with Int]]
 required: Seq[A[X]]
              bar(Seq(new AI, new AS))
                     ^

scala>
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1 回答 1

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不同之处在于 infoo你可能不引用 type X，而 inbar你可以：

// fails
def foo(as: Seq[A[X]] forSome { type X }) = Set.empty[X]

// btw the same:
def foo(as: Seq[A[_]]) = Set.empty[???]  // <-- what would you put here?

// OK
def bar[X](as: Seq[A[X]]) = Set.empty[X]
于 2013-04-25T23:00:09.263 回答