list 列表构造/连接的语法
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Syntax for list construction / concatenation
提问by Colin Woodbury
I've only been at Haskell for two days now, and was wondering what the difference between the two function definitions below are:
我现在只在 Haskell 呆了两天，想知道下面两个函数定义之间的区别是什么：
Prelude> let swap (x1:x2:xs) = x2:x1:xs
Prelude> swap [1..5]
[2,1,3,4,5]
Prelude> let swap' (x1:x2:xs) = [x2] ++ [x1] ++ xs
Prelude> swap' [1..5]
[2,1,3,4,5]
That is, what makes x2:x1:xs different from [x2] ++ [x1] ++ xs ? Please and thanks.
也就是说，是什么使 x2:x1:xs 与 [x2] ++ [x1] ++ xs 不同？请和谢谢。
回答by Jeff Foster
The type signatures are a good place to start:
类型签名是一个很好的起点：
(:) :: a -> [a] -> [a]
(++) :: [a] -> [a] -> [a]
You can find these out with :type (:)and :type (++)in ghci.
您可以找到这些出同:type (:)和:type (++)在ghci中。
As you can see from the type signatures, both are used to produce lists.
从类型签名中可以看出，两者都用于生成列表。
The :operator is used to construct lists (and to take them apart again for pattern matching). To make a list [1,2,3]you just build it up with 1 : 2 : 3 : []. The first element of :is the item to add on the front of the list, and the second element is either a list (also built up with :or the empty list signified by []).
该:运算符用于构造列表（并再次将它们分开以进行模式匹配）。要制作一个列表，[1,2,3]您只需使用1 : 2 : 3 : []. 的第一个元素:是要添加到列表前面的项目，第二个元素是一个列表（也由 构建:或由 表示的空列表[]）。
The ++operator is list concatenation. It takes two lists and appends them together. [1,2,3] ++ [4,5,6]is legal, whereas 1 ++ [1,2,3]is not.
该++运营商连接列表。它需要两个列表并将它们附加在一起。 [1,2,3] ++ [4,5,6]是合法的，而1 ++ [1,2,3]不是。
回答by Ingo
This has nothing to do with syntax. (:) and (++) are just different operators. (:) is a constructor who constructs a list from an element and another list. (++) makes a new list that is the concatenation of two lists. Because (++) is not a constructor you can't use it in patterns.
这与语法无关。(:) 和 (++) 只是不同的运算符。(:) 是一个构造函数，它从一个元素和另一个列表构造一个列表。(++) 生成一个新列表，该列表是两个列表的串联。因为 (++) 不是构造函数，所以不能在模式中使用它。
Now we come to Syntax: the notation
现在我们来到语法：符号
[x2]
that you use is a shorthand for
你使用的是简写
x2:[]
So what you really have done in the second example is:
所以你在第二个例子中真正做的是：
(x2:[]) ++ (x1:[]) ++ xs
Therefore, when constructing a list, you can't avoid (:), it's ultimatively the only way to do it. Note that you must construct intermediate lists to be able to use (++).
因此，在构建列表时，您无法避免 (:)，它最终是唯一的方法。请注意，您必须构造中间列表才能使用 (++)。
