Question

我试图将一个打字版本的React作为Ocaml中的练习。为了使它更具功能性，我将一条记录作为参数传递给渲染。

type ('props,'state) reactInstance =
  {
  props: 'props;
  state: 'state;
  updater: 'a . ('props,'state) reactInstance -> 'a -> 'state -> unit;}
and ('props,'state) reactClass =
  {
  getInitialState: unit -> 'state;
  render: ('props,'state) reactInstance -> element;}

module type ComponentBluePrint  =
  sig
    type props
    type state
    val getInitialState : unit -> state
    val render : (props,state) reactInstance -> element
  end

module type ReactClass  =
  sig
    type props
    type state
    val mkUpdater :
  props ->
    ((props,state) reactInstance -> 'e -> state) ->
      (props,state) reactInstance -> 'e -> unit
    val reactClass : (props,state) reactClass
  end

module CreateComponent(M:ComponentBluePrint) =
  (struct
     include M
     let rec mkUpdater props f i e =
       let nextState = f i e in
       let newInstance =
         { props; state = nextState; updater = (mkUpdater props) } in
       ()

     let reactClass =
       { render = M.render; getInitialState = M.getInitialState }
   end : (ReactClass with type  props =  M.props and type  state =  M.state))

我不明白的一件事是编译器无法推断updater = (mkUpdater props)中let newInstance = { props; state = nextState; updater = (mkUpdater props) }的类型。

Error: Signature mismatch:
       Values do not match:
         let mkUpdater :
  props =>
  (reactInstance props state => '_a => '_b) =>
  reactInstance props state => '_a => unit
       is not included in
         let mkUpdater :
  props =>
  (reactInstance props state => 'e => state) =>
  reactInstance props state => 'e => unit

＆＃39; _a和＆＃39;之间的区别是什么？它看起来与我完全相同。如何进行此类检查？

Answer 1

类型变量'_a（实际字母无关紧要，关键是下划线）是一个所谓的弱类型变量。这是一个不能概括的变量，即它只能用一种具体的类型代替。它就像一个可变的值，但在类型的领域。

用弱类型变量'_a表示的类型不包含在用泛型类型变量表示的类型中。而且，它甚至无法逃脱编译单元，应该隐藏或具体化。

当表达式不是纯值（在语法上定义）时，会创建弱类型变量。通常，它是功能应用程序或抽象。当您通过枚举所有函数参数（即updater = (fun props f i e -> mkUpdater props f i e)）将部分应用的函数替换为正常函数应用程序时，通常可以通过执行所谓的eta-expansion来消除弱类型变量。

Ocaml不正确的类型推断

1 个答案: