Haskell 中的嵌套列表推导式

Question

我正在关注这个Haskell 教程并且在高阶函数部分。它定义了一个名为 chain 的函数：

chain :: (Integral a) => a -> [a]
chain 1 = [1]
chain n
    | even n = n:chain (n `div` 2)
    | odd n = n:chain (n * 3 + 1)

有一个练习可以找到长度超过 15 的“链”的数量。他们这样做：

numLongChains :: Int  
numLongChains = length (filter isLong (map chain [1..100]))  
    where isLong xs = length xs > 15

我试图提出一个列表理解，而不是给我链的数量，而是给我一个来自 [1..100] 的比 15 长的链列表。到目前为止，我最接近的尝试如下：

[ [ a | a <- chain b, length a > 15] | b <- [1..100]]

但我得到：

<interactive>:9:14:
No instance for (Integral [a0]) arising from a use of `chain'
Possible fix: add an instance declaration for (Integral [a0])
In the expression: chain b
In a stmt of a list comprehension: a <- chain b
In the expression: [a | a <- chain b, length a > 15]

<interactive>:9:45:
    No instance for (Enum [a0])
      arising from the arithmetic sequence `1 .. 100'
    Possible fix: add an instance declaration for (Enum [a0])
    In the expression: [1 .. 100]
    In a stmt of a list comprehension: b <- [1 .. 100]
    In the expression:
      [[a | a <- chain b, length a > 15] | b <- [1 .. 100]]

<interactive>:9:46:
    No instance for (Num [a0]) arising from the literal `1'
    Possible fix: add an instance declaration for (Num [a0])
    In the expression: 1
    In the expression: [1 .. 100]
    In a stmt of a list comprehension: b <- [1 .. 100]

我什至接近吗？尽管可能有更好的方法来解决这个问题，但为了学习，我确实想使用嵌套理解来解决这个问题。

Answer 1

You're close. You're looking for:

[ a | b <- [1..10], let a = chain b, length a > 15 ]

The expression

[ [ a | a <- chain b, length a > 15] | b <- [1..100]]

has type:

Integral [a] => [[[a]]]

which is clearly not what you want, but because of Haskell's polymorphic numeric literals, it could possibly type check if the correct definition were in place.

In this case b is inferred to be a list of some type, and that is why you see the error:

No instance for (Integral [a0]) ...

Here is a complete run down on how the types are inferred.

1. from b <- [1..100] we can infer Enum b is a constraint
2. from the call chain b we can infer Integral b is a constraint
3. from a <- chain b we can infer a and b have the same type
4. from length a we can infer a is a list of something, e.g. a ~ [t]

因此，从 (3) 和 (4) 我们可以推断出，b ~ [t]因此我们需要Integral [t]和Enum [t]对于某种类型t。

为了进一步详细说明（3），我们知道chain b是一个列表，例如的类型[t]在哪里。从我们知道与 的元素具有相同的类型，即 的类型也是。tba <- chain bachain bat