Question

我正在学习f＃编译器的内部结构（https://github.com/Microsoft/visualfsharp回购，dev16.0分支），并且偶然发现了其“ typecheked”抽象语法树的有趣部分。简而言之，Expr类型用于表示树的某些节点，这是其定义的一部分：

Expr =
    /// A constant expression. 
    | Const of Const * range * TType

    ///....

    /// Typechecking residue: Indicates a free choice of typars that arises due to 
    /// minimization of polymorphism at let-rec bindings.  These are 
    /// resolved to a concrete instantiation on subsequent rewrites. 
    | TyChoose of Typars * Expr * range

    /// Typechecking residue: A Expr.Link occurs for every use of a recursively bound variable. While type-checking 
    /// the recursive bindings a dummy expression is stored in the mutable reference cell. 
    /// After type checking the bindings this is replaced by a use of the variable, perhaps at an 
    /// appropriate type instantiation. These are immediately eliminated on subsequent rewrites. 
    | Link of Expr ref

TyChoose和Link节点是什么意思？评论提供了一些解释，但对我来说仍然太模糊。是否有任何代码可以澄清此类表达？基本上，我认为类型化的ast不能有一些类型推断算法的残差（换句话说，它仅在将AST转换为带签名的AST的阶段使用）。 Final ast仅具有完整类型的表达式。

UPD ：在转换TAST-> ILX期间似乎未使用Link节点。这是一段代码，表明它已被剥离：

let rec GenExpr (cenv:cenv) (cgbuf:CodeGenBuffer) eenv sp expr sequel =
    let expr =  stripExpr expr
    //......

let rec stripExpr e = 
    match e with 
    | Expr.Link eref -> stripExpr !eref
    | _ -> e

F＃TAST中类型检查残差的节点是什么？

0 个答案: