Existentially quantified types说明: 任何使用小写类型的隐式地与forall关键字开始,所以这两个类型声明在地图是等价的,因为低于声明: id :: a -> a
id :: forall a . a -> a
鉴于斯卡拉的scala.Predef#identity,是否有一个相当于forall,即按照Haskell的上述第二个函数?
为什么我有这个错误? Error:(5, 18) ambiguous reference to overloaded definition, both method startsWith in class String of type (x$1: String)Boolean and method startsWith in class String of type (x$1: String, x