在Antlr4中,如何截获所有INT / NEWLINE令牌解析?
我想要一个听众。
给定语言:

 grammar Expr;
    prog:   (expr NEWLINE)* ;
    expr:   expr ('*'|'/') expr
        |   expr ('+'|'-') expr
        |   INT
        |   '(' expr ')'
        ;
    NEWLINE : [\r\n]+ ;
    INT     : [0-9]+ ;


Css3Lexer lexer = new Css3Lexer(this.file.getStreamAntlr());
Css3Parser parser = new Css3Parser(new CommonTokenStream(lexer));
parser.addParseListener(..)// doesnt resolve.

最佳答案

首先,我将注释您的语法以使侦听器事件更具体:

grammar Expr;
prog: stmt+ EOF;
stmt: expr NEWLINE+;
expr:
    expr ('*'|'/') expr     # Mult
|   expr ('+'|'-') expr     # Add
|   INT                     # Int
|   '(' expr ')'            # Paren
;

NEWLINE : '\r\n';
INT     : [0-9]+ ;


这将允许您创建特定的侦听器,例如:

public class ExprListener : ExprBaseListener
{
    private Stack <int> stack = new Stack <int>();

    public override void ExitInt(ExprParser.IntContext context)
    {
        int i = Convert.ToInt32(context.INT().GetText());
        stack.Push(i);
    }

    public override void ExitStmt(ExprParser.StmtContext context)
    {
        int result = stack.Pop();
        Console.WriteLine("result " + result);
    }

    public override void ExitMult(ExprParser.MultContext context)
    {
        int r = stack.Pop();
        int l = stack.Pop();
        string op = context.GetChild(1).GetText();
        int result;
        if (op == "*")
            result = l * r;
        else
            result = l / r;
        stack.Push(result);
    }

    public override void ExitAdd(ExprParser.AddContext context)
    {
        int r = stack.Pop();
        int l = stack.Pop();
        string op = context.GetChild(1).GetText();
        int result;
        if (op == "+")
            result = l + r;
        else
            result = l - r;
        stack.Push(result);
    }
}

07-24 09:47
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