SML：运算符 ```::``` 是正确结合的并且应该用作 ```<value of type a>::<list with elements of type a>``` 是真的吗？

Question

好吧，我对 cons 运算符有点困惑：:: 我知道这是可以接受的：

   > [2]
   or this
   2::[2,4]
   > [2,2,4]
and i know this is forbbiden
   [2]::[3]
it won't work.
but I recently found out that :
  [2]::[3]::nil
will work :
 > [[2],[3]]
or 
 [2]::[] will do as well.
 >[[2]]

我的问题是为什么？我在想，也许这是一条规则，比如如果我有一个类型为 a 的列表，那么我只能创建一个类型为 a 的新列表。例如，当我有

[2] :: [3]
My lists have elements of type int but I am asking to concatenate into an int list a non init value [3] ??
But when I have [2]::[3]::[] , the empty list [] could be an empty list of any type so here I need it to be an empty list, with elements like [3], so it adds to the nil the element [3] and then the element [2] and so on ...
So is that the reason? Is there something else?

Answer 1

在SML中，列表定义如下：

datatype 'a list = nil | :: of 'a * 'a list

特别地，:: 是类型 'a * 'a list -> 'a list 的中缀构造函数。

语法 [1, 2, 3] 是 1 :: 2 :: 3 :: nil 的糖分。

您不仅需要构造包含 int 的列表（即，其中 'a = int）；例如，你可以有列表 containing int lists，即 int list list。那是你创造的：

[2] :: [3] :: nil
(* can be sugared as *)
[[2], [3]]

（同样，您可以选择 'a = string 并创建列表 "hello" :: "world" :: nil，即 ["hello", "world"]。）

如果要追加两个列表，可以使用中缀 @，其类型为 'a list * 'a list -> 'a list。例如：

[1, 2] @ [3]
(* evaluates to [1, 2, 3] *)

SML：运算符 ```::``` 是正确结合的并且应该用作 ```<value of type a>::<list with elements of type a>``` 是真的吗？

SML: Operator ```::``` is it true that is right assosiative and shoud be used as ```<value of type a>::<list with elements of type a>```?

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