使用不带标签的Core.Std.List.fold_left

问题描述

我正在试验Core的List.fold_left.

I am experimenting with Core's List.fold_left.

# List.fold_left;;
- : 'a Core.Std.List.t -> init:'b -> f:('b -> 'a -> 'b) -> 'b = <fun>

当我指定标签时，效果很好:

It works fine when I specify the labels:

# List.fold_left [1;2;3] ~init:0 ~f:(+);;
- : int = 6

但是当我不指定标签时，我得到了不同的结果:

But I get a different result when I do not specify the labels:

# List.fold_left [1;2;3] 0 (+);;
- : init:(int -> (int -> int -> int) -> '_a) ->
    f:((int -> (int -> int -> int) -> '_a) ->
       int -> int -> (int -> int -> int) -> '_a) ->
    '_a
= <fun>

和其他部分应用程序也产生非直观类型.为什么我可以在list参数后面添加任意数量的0?

And other partial applications also yield nonintuitive types. Why can I add an arbitrary amount of 0's after the list argument?

# List.fold_left [1;2;3] 0;;
- : init:(int -> '_a) -> f:((int -> '_a) -> int -> int -> '_a) -> '_a = <fun>

# List.fold_left [1;2;3] 0 0;;
- : init:(int -> int -> '_a) ->
    f:((int -> int -> '_a) -> int -> int -> int -> '_a) -> '_a
= <fun>

某些其他功能表现出相同的行为:

Some other functions exhibit the same behavior:

# let app ~f ~x = f x;;
val app : f:('a -> 'b) -> x:'a -> 'b = <fun>

# app (fun x -> x + 1) 1;;
- : f:('a -> (int -> int) -> int -> 'b) -> x:'a -> 'b = <fun>

但是，某些带标签的函数在不带标签的情况下应用时可以返回预期的结果.例如:

But some labeled functions are able to return the expected result when applied without labels given. For example:

# List.map;;
- : 'a Core.Std.List.t -> f:('a -> 'b) -> 'b Core.Std.List.t = <fun>

# List.map [1;2;3] (fun x -> x + 1);;
- : int Core.Std.List.t = [2; 3; 4]

为什么某些函数在未指定标签的情况下应用时会返回非直观值，而另一些函数却按预期工作呢?

Why does some functions return nonintuitive values when applied without labels given, while the others work as expected?

推荐答案

根据手册，您仅可以省略标签如果一个应用程序是总计的(省略所有可选参数)"和一个重要的警告，其结果类型为类型变量的ListLabels.fold_left之类的函数将永远不会被认为完全适用."

According to the manual you may only omit labels "if an application is total (omitting all optional arguments)" with the important caveat "that functions like ListLabels.fold_left whose result type is a type variable will never be considered as totally applied."

因此，由于core的fold_left的结果类型当然也是类型变量，因此没有标签就无法调用它.当您执行此操作时，将发生的情况是位置参数被解释为fold_left生成的函数的参数，并且仍在等待您提供带标签的参数.

So since core's fold_left's result type is, of course, also a type variable, you can not call it without labels. What happens when you do is that the positional arguments are interpreted as arguments to the function produced by fold_left and it still waits for you to provide the labeled arguments.

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