英文:
Haskell Type Problem with the ":" operator
问题
-
[]:[[[1]],[]]
-
[[],[1]]:[[]]
The problem is that GHCI gives out:
-
[[],[[1]],[]]
-
[[[],[1]],[]]
Now that doesn't make any sense to me, because I combine lists which do not contain just one kind of element like in 1, or combine two lists with the same type. I need in the front of ":" a non-list item like I wrote down in the top of this question, so why does this work?
英文:
I have to learn Haskell and one exercise was to say If a piece of Code would Work for the Compiler.
The ":" works as a -> [a] -> [a] with a some variable type
My Problem is now the following:
1. []:[[[1]],[]]
And
2. [[],[1]]:[[]]
The Probleme is that GHCI gives out:
1. [[],[[1]],[]]
And
2.[[[],[1]],[]]
Now that doesn't make any Sense to me ,because i combine Lists which dies Not cotaim Just one Kind of Element Like in 1. or in combine two Lists with the Same Type while i need in the Front of ":" a non list Item Like I wrote down in the top of this Question, so why does this Work?
答案1
得分: 3
非空括号只是对(:)链的语法糖。
有两种创建列表的方法。要创建一个空列表，您可以使用[]。要从现有对象列表y和与y元素相同类型的另一个值x创建列表，您可以使用x:y。
因此，括号的任何其他用法都可以转化为空括号和(:)的组合。一些简单的例子：
[a] == a:[][a,b] == a:b:[][a,b,c] == a:b:c:[]
在上述每种情况下，如果a, b, c :: t，那么[] : [t]。
在您的情况下，您有列表的列表，所以开始变得有点混乱，因为无论类型变得多么“嵌套”，[]仍然是空列表。所以对于#1，[]:[[[1]],[]]，您有：
[]：类型为Num a => [[a]]的空列表[[[1]],[]]：解糖为[[1]]:[]:[]的列表，由两个类型为Num a => [[a]]的列表组成，即[[1]]和[]。
因此，表达式的结果实际上是[[],[[1]],[]]或[]:[[1]]:[]:[]。
还有一个令人困惑的地方：在完全解糖的表达式[]:[[1]]:[]:[]中，前两个[]的类型与[[1]]一样，都是Num a => [[a]]。然而，第三个，作为最后一个(:)构造函数的右操作数，具有类型Num a => [[[a]]]。
英文:
Non-empty brackets are just syntactic sugar for chains of (:).
There are two ways to create a list. To create an empty list, you use []. To create a list from an existing list of objects y and another value x of the same type as the elements of y, you use x:y.
So any other use of brackets can be desugared to a combination of empty brackets and (:). Some simple examples:
[a] == a:[][a,b] == a:b:[][a,b,c] == a:b:c:[]
In each of the above cases, if a, b, c :: t, then [] : [t].
In your case, you have lists of lists, so it starts getting a little confusing, because no matter how "nested" the type gets, [] still the empty list. So with #1, []:[[[1]],[]], you have:
[]: the empty list of typeNum a => [[a]][[[1]],[]]: a list that desugars to[[1]]:[]:[], consisting of two lists of typeNum a => [[a]], namely[[1]]and[].
The result of the expression, then, is in fact [[],[[1]],[]] or []:[[1]]:[]:[].
One other confusing point: in the completely desugared expression []:[[1]]:[]:[], the first two occurrences of [] have type Num a => [[a]], just like [[1]]. The third one, though, being the right-hand operand of the last (:) constructor, has type Num a => [[[a]]].
答案2
得分: 0
以下是翻译好的部分：
也许类型提示将有助于说明发生了什么。
对于问题1：
([] :: [[Integer]])
：
[([[1]] :: [[Integer]])
,([] :: [[Integer]])
]
对于问题2：
([([] :: [Integer])
,([1] :: [Integer])
] :: [[Integer]]
)
：
([[]] :: [[[Integer]]])
你可以将这些示例粘贴到 REPL 中，以查看它们能够正常工作。
空列表的类型是 forall a. [a]，其中 a 可以是任何类型，包括 [[Integer]]，这就是为什么这些示例能够正常工作的原因。
英文:
Perhaps type hints will help illustrate what's going on.
For problem 1:
([] :: [[Integer]])
:
[ ([[1]] :: [[Integer]])
, ([] :: [[Integer]])
]
For problem 2:
([ ([] :: [Integer])
, ([1] :: [Integer])
] :: [[Integer]]
)
:
([[]] :: [[[Integer]]])
You should be able to paste each of those examples into a REPL to see that they work.
An empty List is of type forall a. [a] where a can be anything, including [[Integer]], which is why these examples work.
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