是否可以在F＃的同一行中的一个类中编写几个成员？

Question

我目前有以下代码：

let private matchers = [|
    { new Matcher() with
        member __.RegexExpr = RegexExpr("^(let)")
        member __.TokenTypeCreator _ = TokenType.LET }
|]

我想知道是否可以在同一行中写出所有内容。类似于（非工作）的东西：

let private matchers = [|
    { new Matcher() with member __.RegexExpr = RegexExpr("^(let)"); member __.TokenTypeCreator _ = TokenType.LET }
|]

由于

Answer 1

我也找不到这样做的语法。只需在成员之间放置任何内容即可在单行上声明接口：

type IFoo = abstract member A: int abstract member B: int;;

但是如果我尝试相同来实例化它，编译器会拒绝它：

  { new Foo with member this.A = 1 member this.B = 2 };;
  ---------------------------------^^^^^^

stdin(10,34): error FS0010: Unexpected keyword 'member' in expression. Expected incomplete structured construct at or before this point or other token.

我不得不说我有时希望F＃对基于缩进的基于缩进的代码有一个统一的规则，就像Haskell那样。