scala - 二维数组作为函数

Question

可以使用一维数组作为函数

def foo1(f: Int => Int) = ???
foo1(Array(1))

可以使用带有两个参数列表的函数

def foo2(f: Int => Int => Int) = ???
def plus(x: Int)(y: Int) = x + y
foo2(plus)

我可以声明一个接受二维数组的函数吗 Array(Array(1))没有实际使用 Array输入函数声明？还是隐式转换为 Int => Array[Int]就这样？

Answer 1

对于任意嵌套数组，您可以使用类型体操的“深度”隐式转换

  trait ToIdxFunction[X[_], A] {
    type Result
    def toIdx(x: X[A]): Int => Result
  }

  trait LowerPriorityDeepFunctor {
    implicit def plainArray[A] =
      new ToIdxFunction[Array, A] {
        type Result = A
        def toIdx(x: Array[A]): Int => Result = {
          i => x(i)
        }
      }
  }

  object ToIdxFunction extends LowerPriorityDeepFunctor {
    implicit def nestedArray[A](implicit inner: ToIdxFunction[Array, A]) = {
      new ToIdxFunction[Array, Array[A]] {
        type Result = Int => inner.Result
        def toIdx(x: Array[Array[A]]): Int => Result = {
          i => inner.toIdx(x(i))
        }
      }
    }
  }

  import ToIdxFunction._

  implicit class Ops[X[_], A](self: X[A]) {
    def asFunction(implicit F: ToIdxFunction[X, A]) = F.toIdx(self)
  }

Scala 控制台中的示例

scala> Array(1).asFunction
res4: Int => Int = <function1>

scala>   Array(Array(1)).asFunction
res5: Int => (Int => Int) = <function1>

scala>

scala>   Array(Array(Array(1))).asFunction
res6: Int => (Int => (Int => Int)) = <function1>

scala>   Array(Array(Array(Array(1)))).asFunction
res7: Int => (Int => (Int => (Int => Int))) = <function1>

这有效:

  def foo(f: Int => Int => Int => Int) = println(f(0)(0)(0))
  foo(Array(Array(Array(1))).asFunction)