为什么当传入空列表时我的排列函数会发出警告？
我的排列函数：
fun perms [] = [[]]
| perms (x::xs) = let
fun insertEverywhere [] = [[x]]
| insertEverywhere (y::ys) = let
fun consY list = y::list
in
(x::y::ys) :: (map consY (insertEverywhere ys))
end
in
List.concat (map insertEverywhere (perms xs))
end;
输入：
perms [];
输出：
stdIn:813.1-813.9 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val it = [[]] : ?.X1 list list
有人可以解释为什么类型变量没有被泛化吗？
需要注意的是，权限类型是在输入权限后给出的；所以
perms;
val it = fn : 'a list -> 'a list list
看起来我已经实现了广义变量，至少对我来说是这样。
my permutation function:
fun perms [] = [[]]
| perms (x::xs) = let
fun insertEverywhere [] = [[x]]
| insertEverywhere (y::ys) = let
fun consY list = y::list
in
(x::y::ys) :: (map consY (insertEverywhere ys))
end
in
List.concat (map insertEverywhere (perms xs))
end;
input:
perms [];
output:
stdIn:813.1-813.9 Warning: type vars not generalized because of
value restriction are instantiated to dummy types (X1,X2,...)
val it = [[]] : ?.X1 list list
Can someone explain why the type vars aren't generalized?
I should note, the type of perms is given after inputting perms; as
perms;
val it = fn : 'a list -> 'a list list
So it looks like I have achieved generalized variables, to me at least.
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空列表是一个特殊列表，其元素可能具有任何类型。当您调用
perms []时，编译器会对元素的类型感到困惑。您可以使用：或者
编译器会很高兴，因为 in 可以推断列表的特定类型。
Empty list is a special list which could potentially have any type for its elements. When you invoke
perms [], the compiler is confused about the type of elements. You can either use:or
then the compiler is happy because in can inference a specific type of the lists.