如何编写一个 apply 接受多个参数的函数？
这是一个人为的例子：
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
上面给我一个类型推断错误？
Define compose2，例如作为扩展方法Function1:
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
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