使用 ListLike 的不明确类型
我正在 Haskell 中编写一个函数，从任何带有 Ord 元素的 ListLike 制作直方图：
import qualified Data.ListLike as LL
...
frequencies :: (Ord x, LL.ListLike xs x) => xs -> [(x, Int)]
frequencies xs = LL.map (\x->(LL.head x, LL.length x)) $ LL.group $ LL.sort xs
当尝试编译上述代码时，我收到一条有关不明确类型的错误消息：
Ambiguous type variable `full0' in the constraint:
(LL.ListLike full0 xs) arising from a use of `LL.group'
Probable fix: add a type signature that fixes these type variable(s)
In the expression: LL.group
In the second argument of `($)', namely `LL.group $ LL.sort xs'
In the expression:
LL.map (\ x -> (LL.head x, LL.length x)) $ LL.group $ LL.sort xs
< code>LL.group 的类型为 (ListLike full0 full, ListLike full item, Eq item) =>满-> full0 对应于 (Eq a) => [a]->[[a]] 用普通列表来说。
我不明白为什么类型不明确会出现问题。 Haskell 是否无法推断出存在这样一种类型：“ListLike with full as elements”，即 full0？
I'm writing a function in Haskell to make a histogram from any ListLike with Ord elements:
import qualified Data.ListLike as LL
...
frequencies :: (Ord x, LL.ListLike xs x) => xs -> [(x, Int)]
frequencies xs = LL.map (\x->(LL.head x, LL.length x)) $ LL.group $ LL.sort xs
When trying to compile the above code i get an error message about ambiguous types:
Ambiguous type variable `full0' in the constraint:
(LL.ListLike full0 xs) arising from a use of `LL.group'
Probable fix: add a type signature that fixes these type variable(s)
In the expression: LL.group
In the second argument of `($)', namely `LL.group $ LL.sort xs'
In the expression:
LL.map (\ x -> (LL.head x, LL.length x)) $ LL.group $ LL.sort xs
LL.group has type (ListLike full0 full, ListLike full item, Eq item) => full -> full0 which corresponds to (Eq a) => [a]->[[a]] speaking in terms of ordinary lists.
I dont understand why there is a problem with ambigious types. Is Haskell somehow unable to infer that there is such a type as "ListLike with full as elements", i.e full0?
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不，问题是有太多类型可供选择
full0，而 Haskell 编译器不知道该使用哪个。想一想：一般来说，可以有多个类型full0，它是一种类似列表的类型，以full作为元素：[full]或Data.Sequence.Seq full或许多其他选择。并且由于LL.map定义的通用性，full0不被限制为[full]来自函数频率的返回类型。如果要将
full0限制为[full]，一种方法是替换定义中的与普通的旧LL.map频率map列表。No, the problem is there are too many types to choose
full0from and the Haskell compiler does not know which to use. Think about it: in general, there can be more than one typefull0which is a list-like type withfullas elements:[full]orData.Sequence.Seq fullor many other choices. And because of the generality of the definition ofLL.map,full0is not constrained to be[full]from the return type of the functionfrequencies.If you want to restrict
full0to[full], one way to do this is to replaceLL.mapin the definition offrequencieswith the plain oldmapfor lists.