haskell - 如何评估 `fix f = let {x = f x} in x`？

Question

fix f = let {x = f x} in x谈let , 我以为 let P = Q in R将评估 Q -> Q' 然后 P 被 R 中的 Q' 替换，或者:R[P -> Q'] .
但是在 fix定义Q取决于R，那么如何评估呢？
我想这是关于惰性评估的。 Q'变成了一个重击，但我无法在我的脑海中推理。
作为上下文，我正在查看 是 组合器，它应该找到一个函数的不动点，所以如果我有这个函数，one x = 1 ，然后 fix one == 1必须坚持，对吧？
所以fix one = let {x = one x} in x ，但我看不到 1会从中出现。

Answer 1

Talking about let, I thought that let P = Q in R would evaluate Q -> Q' then P is replaced by Q' in R, or: R[P -> Q'].

在道德上，是的，但是 P不会立即评估，而是在需要时评估。

But in fix definition the Q depends on R, how to evaluate then?

Q不依赖于 R , 这取决于 P .这使得 P递归地依赖于自身。这可能导致几种不同的结果。粗略地说:

如果 Q在评估 P 之前不能返回其结果的任何部分，然后 P表示无限递归的计算，它不会终止。例如，

let x = x + 1 in x     -- loops forever with no result
-- (GHC is able to catch this specific case and raise an exception instead,
-- but it's an irrelevant detail)

如果 Q可以在需要评估 P 之前返回其结果的一部分，它确实如此。

let x = 2 : x in x
-- x = 2 : .... can be generated immediately
-- This results in the infinite list 2:2:2:2:2:.....

let x = (32, 10 + fst x) in x
-- x = (32, ...) can be generated immediately
-- hence x = (32, 10 + fst (32, ...)) = (32, 10+32) = (32, 42)

I imagine that this is about lazy evaluation. Q' becomes a thunk, but I can't reason this in my head.

P与 thunk 相关联。重要的是这个 thunk 在返回输出的某些部分之前是否调用自己。

As a matter of context, I'm looking at Y combinator, it should find a fixed point of a function so if I have this function. one x = 1, then fix one == 1 must hold, right?

是的。

So fix one = let x = one x in x, but I can't see why 1 would emerge from that

我们可以这样计算:

fix one
 = {- definition of fix -}
let x = one x in x
 = {- definition of x -}
let x = one x in one x
 = {- definition of one -}
let x = one x in 1
 = {- x is now irrelevant -}
1

只需扩展定义。保留递归定义，以防您再次需要它们。