scala - 组合具有多个参数的函数

Question

如何编写一个 apply 接受多个参数的函数？

这是一个人为的示例:

val sum: List[Int] => Int = l => l.sum

val double: Int => Int = i => i * i

double.compose(sum).apply(List(1,2,3)) //=> 36

val sumAppend: (List[Int], Int) => Int = (l, i) => i :: l sum

double.compose(sumAppend).apply(List(1,2,3), 1) // Attempt to append 1 to list then sum

上面给了我一个类型推断错误？

Answer 1

定义compose2，例如作为Function1的扩展方法:

implicit class ComposeFunction2[A, B, C, D](f1: Function1[C, D]) {
  def compose2(f2: Function2[A, B, C]): Function2[A, B, D] =
    (a: A, b: B) => f1(f2(a, b))
}

这将比替代方案更快，因为它不分配元组。用法:

scala> val double: Int => Int = i => i * i
double: Int => Int = <function1>

scala> val sumAppend: (List[Int], Int) => Int = (l, i) => i :: l sum
sumAppend: (List[Int], Int) => Int = <function2>

scala> double.compose2(sumAppend).apply(List(1,2,3), 1)
res5: Int = 49