Haskell:为什么((...(.))f g等于f. g x? [英] Haskell: Why is ((.).(.)) f g equal to f . g x?

问题描述

能否请您解释表达式((...(.)))的含义? 据我所知(.)的类型为(b-> c)->(a-> b)-> a-> c.

Could you please explain the meaning of the expression ((.).(.))? As far as I know (.) has the type (b -> c) -> (a -> b) -> a -> c.

推荐答案

(.) . (.)是组合运算符及其本身的组合.

(.) . (.) is the composition of the composition operator with itself.

如果我们看

((.) . (.)) f g x

我们可以评估几个步骤，首先我们将其括起来，

we can evaluate that a few steps, first we parenthesise,

((((.) . (.)) f) g) x

然后我们使用(foo . bar) arg = foo (bar arg)进行申请:

then we apply, using (foo . bar) arg = foo (bar arg):

~> (((.) ((.) f)) g) x
~> (((.) f) . g) x
~> ((.) f) (g x)
~> f . g x

更多原则

(.) :: (b -> c) -> (a -> b) -> (a -> c)

因此，使用(.)作为(.)的第一个参数，我们必须统一

So, using (.) as the first argument of (.), we must unify

b -> c

使用

(v -> w) -> (u -> v) -> (u -> w)

这产生了

b = v -> w
c = (u -> v) -> (u -> w)

和

(.) (.) = ((.) .) :: (a -> v -> w) -> a -> (u -> v) -> (u -> w)

现在，要将其应用于(.)，我们必须统一类型

Now, to apply that to (.), we must unify the type

a -> v -> w

重命名后

类型为(.)的

with the type of (.), after renaming

(s -> t) -> (r -> s) -> (r -> t)

产生

a = s -> t
v = r -> s
w = r -> t

因此

(.) . (.) :: (s -> t) -> (u -> r -> s) -> (u -> r -> t)

从类型中，我们可以(几乎)看到(.) . (.)将一个参数的函数应用于两个参数的函数的结果.

and from the type we can (almost) read that (.) . (.) applies a function (of one argument) to the result of a function of two arguments.

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