C#中的解析器并打印AST [英] Parser in C# and printing AST

问题描述

我正在用C#实现复杂语法的AST(抽象语法树)，但是，为了使这个问题更简单，我将使用非常简单的语法.

I am implementing an AST (Abstract Syntax Tree) in C# for a complex grammar, however, to make this question simple, I will use a very simple grammar.

考虑以下语法:

rules Expr ::= Term "+" Term 
        | Term ;

rules Term ::= Ident
        | Integer ;

我使用了bnfc并生成了解析器/词法分析器，直指我可以解析一段代码并可以打印解析树.现在，我想将其映射到AST，并打印抽象语法树". 此处是我到目前为止在一个示例项目中所做的.

I have used bnfc and generated the parser/lexer and got to the point that I can parse a piece of code and can print the parse tree. Now I want to map it to AST, and print the Abstract Syntax Tree. here is what I have done so far in a sample project.

但是，目前，当我测试程序时，我的AST返回为NULL.

However, currently when I test the program, my AST comes back as NULL.

var astGen = new gplex.VisitSkeleton.ExprVisitor<Expr1, gplex.Absyn.Expr1>();
var ast = astGen.Visit((gplex.Absyn.Expr1)parse_tree, (gplex.Absyn.Expr1)parse_tree); 

在这里，ast为空.拥有C#经验的人可以帮助我做到这一点吗?

Here, ast is null. Can someone with experience in C# help me get on track with this?

推荐答案

在您的项目中，我注意到您所有的通用Visit方法最终都会返回相同的常量结果，无论这些方法的其余主体在此之前做了什么.

At your project, I noticed that all your generic Visit methods end up returning the same constant result no matter what the rest of those methods' bodies has done prior to that:

return default(R);

对于作为引用类型的R的具体类型，此"default(R)"返回值的确将始终为空.

For concrete types of R that are reference types, this "default(R)" return value will indeed be always null.

'HTH，

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