中止分析错误并显示有用消息

问：

我有一个 ANTLR 4 语法，并从中构建了一个词法分析器和解析器。现在，我正在尝试以这样一种方式实例化该解析器，以便它将解析直到遇到错误。如果遇到错误，则不应继续分析，但应提供有关问题的有用信息;理想情况下，机器可读的位置和人类可读的消息。

这是我目前所拥有的：

grammar Toy;

@parser::members {

    public static void main(String[] args) {
        for (String arg: args)
            System.out.println(arg + " => " + parse(arg));
    }

    public static String parse(String code) {
        ErrorListener errorListener = new ErrorListener();
        CharStream cstream = new ANTLRInputStream(code);
        ToyLexer lexer = new ToyLexer(cstream);
        lexer.removeErrorListeners();
        lexer.addErrorListener(errorListener);
        TokenStream tstream = new CommonTokenStream(lexer);
        ToyParser parser = new ToyParser(tstream);
        parser.removeErrorListeners();
        parser.addErrorListener(errorListener);
        parser.setErrorHandler(new BailErrorStrategy());
        try {
            String res = parser.top().str;
            if (errorListener.message != null)
                return "Parsed, but " + errorListener.toString();
            return res;
        } catch (ParseCancellationException e) {
            if (errorListener.message != null)
                return "Failed, because " + errorListener.toString();
            throw e;
        }
    }

    static class ErrorListener extends BaseErrorListener {

        String message = null;
        int start = -2, stop = -2, line = -2;

        @Override
        public void syntaxError(Recognizer<?, ?> recognizer,
                                Object offendingSymbol,
                                int line,
                                int charPositionInLine,
                                String msg,
                                RecognitionException e) {
            if (message != null) return;
            if (offendingSymbol instanceof Token) {
                Token t = (Token) offendingSymbol;
                start = t.getStartIndex();
                stop = t.getStopIndex();
            } else if (recognizer instanceof ToyLexer) {
                ToyLexer lexer = (ToyLexer)recognizer;
                start = lexer._tokenStartCharIndex;
                stop = lexer._input.index();
            }
            this.line = line;
            message = msg;
        }

        @Override public String toString() {
            return start + "-" + stop + " l." + line + ": " + message;
        }
    }

}

top returns [String str]: e* EOF {$str = "All went well.";};
e: 'a' 'b' | 'a' 'c' e;

将此内容保存到 ，然后尝试以下命令：Toy.g

> java -jar antlr-4.5.2-complete.jar Toy.g
> javac -cp antlr-4.5.2-complete.jar Toy*.java
> java -cp .:tools/antlr-4.5.2-complete.jar ToyParser ab acab acc axb abc
ab => All went well.
acab => All went well.
acc => Failed, because 2-2 l.1: no viable alternative at input 'c'
axb => Parsed, but 1-1 l.1: token recognition error at: 'x'
Exception in thread "main" org.antlr.v4.runtime.misc.ParseCancellationException
    at org.antlr.v4.runtime.BailErrorStrategy.recoverInline(BailErrorStrategy.java:90)
    at org.antlr.v4.runtime.Parser.match(Parser.java:229)
    at ToyParser.top(ToyParser.java:187)
    at ToyParser.parse(ToyParser.java:95)
    at ToyParser.main(ToyParser.java:80)
Caused by: org.antlr.v4.runtime.InputMismatchException
    at org.antlr.v4.runtime.BailErrorStrategy.recoverInline(BailErrorStrategy.java:85)
    ... 4 more

一方面，我觉得我已经做得太多了。看看我为一个简单而常见的任务写了多少代码，我不禁怀疑我是否错过了一些更简单的解决方案。另一方面，即使这样似乎还不够，原因有两个。首先，虽然我设法报告了词法分析器错误，但它们仍然无法阻止解析器继续处理剩余的流。这是输入字符串的证据。其次，我仍然有错误没有被报告给错误侦听器，正如堆栈跟踪所证明的那样。Parsed, butaxb

如果我不安装 ，我会得到更有用的输出：BailErrorStrategy

acc => Parsed, but 2-2 l.1: mismatched input 'c' expecting 'a'
axb => Parsed, but 1-1 l.1: token recognition error at: 'x'
abc => Parsed, but 2-2 l.1: extraneous input 'c' expecting {<EOF>, 'a'}

有没有办法获得这种错误消息，但仍然因错误而保释？我从消息来源可以看出，该消息确实是由 生成的，显然是在它弄清楚如何解决问题之后。我是否应该让它这样做，然后保释，即在投掷之前编写我自己的变体调用 super？extraneous inputDefaultErrorStrategyBailErrorStrategy