Haskell - 柯里？需要进一步解释

问题描述

  addList :: [int]  - > int 
 addList = foldl1（+）
  

为什么这会起作用？ Currying部分。为什么没有变量？

解决方案

如果你定义了一个像 fxy = bla ，这与 fx = \y - >相同。 bla ，它与 f = \ x - >相同。 （\y-> bla）。换句话说， f 是一个函数，它接受一个参数 x ，并返回另一个带有一个参数的函数 y ，然后返回实际结果。这就是所谓的currying。

类似地，当您执行 fxy 时，它与 （fx）y 。即你使用参数 x 调用函数 f 。这将返回另一个函数，您将它应用于参数 y 。

换句话说，当您执行 addList xs = foldl1（+）xs ，您首先调用 foldl1（+），然后返回另一个函数，适用于 xs 。因此，由于 foldl1（+）返回的函数实际上与 addList 相同，所以您可以将其缩短为 addList = foldl1（+）。

So something like

addList :: [int] -> int
addList = foldl1 (+)

Why does this work? The Currying part. Why no variable?

解决方案

If you define a function like f x y = bla, this is the same as f x = \y -> bla, which is the same as f = \x -> (\y -> bla). In other words f is a function which takes one argument, x, and returns another function which takes one argument, y, and then returns the actual result. This is known as currying.

Analogously when you do f x y, it's the same as (f x) y. I.e. you're calling the function f with the argument x. This returns another function, which you apply to the argument y.

So in other words when you do addList xs = foldl1 (+) xs, you're first calling foldl1 (+) which then returns another function, which you apply to xs. So since the function returned by foldl1 (+) is actually the same one as addList, you can just shorten it to addList = foldl1 (+).

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