“_”类型在快速错误消息中的含义是什么？

举个例子,我得到错误：

Cannot convert value of type 'JawDroppingFeat<Superhero>' to closure result type 'JawDroppingFeat<_>'

当我尝试编译时：

protocol SuperheroType {
  typealias Superpower
}

struct JawDroppingFeat<Superhero: SuperheroType where Superhero: Arbitrary,Superhero.Superpower: Arbitrary>: Arbitrary {
  let subject: Superhero
  let superpowerUsed: Superhero.Superpower

  static var arbitrary: Gen<JawDroppingFeat<Superhero>> {
    get {
      return Gen.zip(Superhero.arbitrary,Superhero.Superpower.arbitrary)
        .map{ (x: Superhero,y: Superhero.Superpower) in
          JawDroppingFeat(subject: x,superpowerUsed: y)
        }
    }
  }
}

Gen和Arbitrary类型来自SwiftCheck,相关声明如下：

public struct Gen<A> {
  public static func zip<A,B>(gen1: SwiftCheck.Gen<A>,_ gen2: SwiftCheck.Gen<B>) -> SwiftCheck.Gen<(A,B)>
  public func map<B>(f: A -> B) -> SwiftCheck.Gen<B>
}

public protocol Arbitrary {
  public static var arbitrary: SwiftCheck.Gen<Self> { get }
}

我假设< _>与迅速未能推断出类型参数而不是Chris Lattner对我畏缩的图像有关.但它有更精确(和记录)的含义吗？

编辑

我当前最好的理论是,当Swift无法推断出一个类型参数,而不是立即失败时,它会分配一个null(_)类型,这导致实际的编译错误在某个不兼容类型的某个点下游(在我的例子中,传递.map的参数).