Scala Rank-1 多态性

Question

以下代码片段来自twitter的Scala学派:

Scala has rank-1 polymorphism. Roughly, this means that there are some type concepts you’d like to express in Scala that are “too generic” for the compiler to understand. Suppose you had some function:

def toList[A](a: A) = List(a)

which you wished to use generically:

def foo[A, B](f: A => List[A], b: B) = f(b)

This does not compile, because all type variables have to be fixed at the invocation site. Even if you “nail down” type B,

def foo[A](f: A => List[A], i: Int) = f(i) // Line 1

…you get a type mismatch.

为什么第 1 行会失败？ B 的类型是已知的。为什么编译失败？

Answer 1

scala> def toList[A](a:A) = List(a)
toList: [A](a: A)List[A]

scala> def foo[A, B](f: A => List[A], b: B) = f(b)
<console>:10: error: type mismatch;
 found   : b.type (with underlying type B)
 required: A
       def foo[A, B](f: A => List[A], b: B) = f(b)

此行无法编译，因为当函数需要 A 类型的值时，您传递的是 B 类型的值。

def foo[A](f: A => List[A], i: Int) = f(i) // Line 1

对于这一行，您需要提供从类型 Int 到类型 A 的隐式转换。