为什么这个函数的类型(a - > a) - >一个? [英] Why is the type of this function (a -> a) -> a?
问题描述
为什么这个函数的类型(a - > a) - > a?
Prelude>让y f = f(y f)
前奏> :t y
y ::(t - > t) - > t
不应该是无限/递归类型吗?
我会试着把它认为应该是的类型,但我不能这样做。
y ::(t - > t) - > ?WTFIsGoingOnOnTheRHS?
我没有得到f(y f)如何解析为一个值。以下对我更有意义:
Prelude>让y f x = f(y f)x
前奏> :(a - > b) - > a - > b) - > a - > b
但它仍然令人无法容易地混淆。怎么回事?
(a - > b) - >对于某些 a
, b
和 c code>我们还不知道;毕竟,它需要一个函数 f
,并将其应用于参数,因此它必须是一个函数。
由于 yf = fx
(同样,对于一些 x
),我们知道返回类型 y
必须是 f
本身的返回类型。因此,我们可以细化一下 y
的类型:它必须是(a - > b) - > b
对于某些 a
和 b
我们还不知道。
为了弄清楚 a
是什么,我们只需要看看传给的值的类型, ˚F
。它是 y f
,这是我们试图找出现在类型的表达式。我们说 y
的类型是(a - > b) - > b
(对于某些 a
, b
等),所以我们可以说这个 yf
的应用程序必须是 b
本身。
<因此, f
的参数类型是 b
。把它们全部放在一起,我们得到(b - > b) - > b
- 当然,它与(a - > a) - >相同。一个
。
以下是一个更直观但不太准确的事物视图:我们说 yf = f (yf)
,我们可以扩展到等价的 yf = f(f(yf))
, yf = f (f(f(yf)))
,依此类推。因此,我们知道我们总是可以在整个事情中应用另一个 f
,并且因为所讨论的整体是应用 f 到一个参数, f
必须有 a - >一个
;并且由于我们只是得出结论,整个事情是将 f
应用于参数的结果,所以返回类型 y
必须是 f
本身 - 再次一起作为(a - > a) - > a
。
Why is the type of this function (a -> a) -> a?
Prelude> let y f = f (y f)
Prelude> :t y
y :: (t -> t) -> t
Shouldn't it be an infinite/recursive type?
I was going to try and put into words what I think it's type should be, but I just can't do it for some reason.
y :: (t -> t) -> ?WTFIsGoingOnOnTheRHS?
I don't get how f (y f) resolves to a value. The following makes a little more sense to me:
Prelude> let y f x = f (y f) x
Prelude> :t y
y :: ((a -> b) -> a -> b) -> a -> b
But it's still ridiculously confusing. What's going on?
解决方案 Well, y
has to be of type (a -> b) -> c
, for some a
, b
and c
we don't know yet; after all, it takes a function, f
, and applies it to an argument, so it must be a function taking a function.
Since y f = f x
(again, for some x
), we know that the return type of y
must be the return type of f
itself. So, we can refine the type of y
a bit: it must be (a -> b) -> b
for some a
and b
we don't know yet.
To figure out what a
is, we just have to look at the type of the value passed to f
. It's y f
, which is the expression we're trying to figure out the type of right now. We're saying that the type of y
is (a -> b) -> b
(for some a
, b
, etc.), so we can say that this application of y f
must be of type b
itself.
So, the type of the argument to f
is b
. Put it all back together, and we get (b -> b) -> b
— which is, of course, the same thing as (a -> a) -> a
.
Here's a more intuitive, but less precise view of things: we're saying that y f = f (y f)
, which we can expand to the equivalent y f = f (f (y f))
, y f = f (f (f (y f)))
, and so on. So, we know that we can always apply another f
around the whole thing, and since the "whole thing" in question is the result of applying f
to an argument, f
has to have the type a -> a
; and since we just concluded that the whole thing is the result of applying f
to an argument, the return type of y
must be that of f
itself — coming together, again, as (a -> a) -> a
.
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