使用有用的消息中止解析错误

Question

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.现在我试图以这样的方式实例化该解析器,它将解析直到遇到错误.如果遇到错误,则不应继续解析,但应提供有关问题的有用信息; 理想情况下是机器可读的位置和人类可读的消息.

这就是我现在所拥有的:

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;

将其保存到Toy.g,然后尝试以下命令:

> 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.看看我写了多少代码应该是一个简单而常见的任务,我不禁想知道我是否缺少一些简单的解决方案.另一方面,由于两个原因,即便这样也不够.首先,当我设法报告lexer错误时,它们仍然不会阻止解析器继续处理剩余的流.这是Parsed, but输入字符串的证据axb.其次,我仍然留下错误,这些错误没有报告给错误监听器,如堆栈跟踪所证明的那样.

如果我不安装BailErrorStrategy,我得到更有用的输出:

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'}

有没有办法得到这种错误信息但仍然保释错误？我可以从消息来源看到这个extraneous input消息确实是由DefaultErrorStrategy它产生的,显然是在它解决了如何修复问题之后.我应该让它做到这一点,然后拯救,即BailErrorStrategy在投掷之前写出我自己的变种哪个叫super？

Answer 1

在同样的情况下,我最终得到了扩展DefaultErrorStrategy和重写report*方法.这很简单(你也可以使用ANTLRErrorStrategy).

在这里,您可以找到失败快速策略的示例.我认为在您的情况下,您可以以相同的方式收集所有错误并构建详细的报告.