如何定义接受柯里化函数参数的函数? [英] How to define function accepting curried function parameter?

问题描述

fn2以下编译失败，

def fn(x: Int)(y: Int) = x + y
def fn2(f: ((Int)(Int)) => Int) = f
fn2(fn)(1)(2) // expected = 3

如何定义fn2来接受fn?

推荐答案

应该是这样的:

scala> def fn2(f: Int => Int => Int) = f
fn2: (f: Int => (Int => Int))Int => (Int => Int)

scala> fn2(fn)(1)(2)
res5: Int = 3

(Int)(Int) =>Int 不正确 - 您应该使用 Int =>整数 =>Int(就像在 Haskell 中一样).实际上，柯里化函数接受 Int 并返回 Int =>Int 函数.

(Int)(Int) => Int is incorrect - you should use Int => Int => Int (like in Haskell), instead. Actually, the curried function takes Int and returns Int => Int function.

附言您也可以使用 fn2(fn _)(1)(2)，因为在前面的示例中传递 fn 只是 eta 扩展的一种简短形式，请参阅 这些scala方法中下划线用法的区别.

P.S. You could also use fn2(fn _)(1)(2), as passing fn in previous example is just a short form of eta-expansion, see The differences between underscore usage in these scala's methods.

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