Scala推断的类型参数 - 类型范围推断为'Nothing' [英] Scala inferred type arguments - Type bounds inferring to 'Nothing'

问题描述

我试图写一个简单的查询monad，并且无法正常使用我的通用类型注释。

我第一次尝试去如下所示（为了简明起见，大大简化了）

pre codease case class Person（val name：String）
抽象类Schema [T ]
对象人们扩展了Schema [Person]

case class Query [U<：Schema [T]，T]（schema：U）{< ---- Type signature
def结果：Seq [T] = ...
def其中（f：U =>操作）= ...
}

类TypeText extends Application {
val query = Query（People）< ---- Type type inference
}

编译器不喜欢这样，因为它不能推断'T'的类型。

错误：推断类型参数[People.type，Nothing]不符合方法apply的类型参数bounds [U<：Schema [T]，T]

hile试验我发现使用视图边界可以按预期工作

  case class Query [U<％Schema [T]，T ]（schema：U）{
  

（注意使用视图边界<％类型绑定<：）

然而，在我对类型系统的有限理解中，因为我期望Schema的实际子类（而不仅仅是可转换性） T]，我会假设类型绑定<是在这里使用的正确界限？

如果是这样的话，我错过了什么 - 怎么做我给编译器足够的提示，以便在使用类型边界而不是视图边界时正确推断T？

在两个类型参数之间，您可以使用类似于以下内容的类：

  case class Query [U，T]（schema：U） ev：U <：< Schema [T]）{...} 
  

请参阅 Scala Language Spec 获取更多信息。

I'm attempting to write a simple query monad and am having trouble getting my generic type annotations correct.

My first attempt went as follows (vastly simplified for conciseness)

case class Person( val name: String )
abstract class Schema[T]    
object People extends Schema[Person]

case class Query[U <: Schema[T], T]( schema: U ) {      <---- Type signature
    def results: Seq[T] = ...
    def where( f: U => Operation ) = ...
}

class TypeText extends Application {
    val query = Query( People )                     <---- Type inference fails
}

The compiler didn't like this, as it couldn't infer the type of 'T'.

error: inferred type arguments [People.type,Nothing] do not conform to method apply's type parameter bounds [U <: Schema[T],T]

While experimenting I found that using view bounds instead works as expected

case class Query[U <% Schema[T], T]( schema: U ) {

(Note the use of view bound "<%" instead of type bound "<:")

However in my limited understanding of the type system, since I'm expecting an actual subclass (and not just convertibility) of Schema[T], I would assume type bound "<:" is the correct bounds to be using here?

If this is the case, what am I missing - how do I give the compiler enough hints to infer T correctly when using type bounds instead of view bounds?

解决方案

In order to encode the relationship between the two type parameters, you can use something like

case class Query[U, T](schema: U)(implicit ev: U <:< Schema[T]) { ... }

See §4.3 and §4.4 of the Scala Language Spec for more info.

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