Question

  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => f(a, _)




  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => (b: B) => f(a, b)

我在想的是，在第一个实现中，我将使用函数f并传入A和任何东西（但编译器将键入检查第二个参数是否为B）以获得C out。

Answer 1

是的，它们完全相同。如果你编译：

object Test {
  def curry[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => f(a, _)

  def curry2[A,B,C](f: (A, B) => C): A => (B => C) =
    (a: A) => (b: B) => f(a, b)
}

使用-Xprint:typer，您将获得中间抽象语法树：

[[syntax trees at end of                     typer]]
package <empty> {
  object Test extends scala.AnyRef {
    def <init>(): Test.type = {
      Test.super.<init>();
      ()
    };
    def curry[A, B, C](f: (A, B) => C): A => (B => C) = ((a: A) => ((x$1: B) => f.apply(a, x$1)));
    def curry2[A, B, C](f: (A, B) => C): A => (B => C) = ((a: A) => ((b: B) => f.apply(a, b)))
  }
}

在＆＃34; typer＆＃34;阶段，当编译器为所有内容分配类型时，它会意识到_（现在名为x$1）必须是B类型。

这两种咖喱功能实现是否相同？

1 个答案: