在antlr4中精确分析n个参数

我将Antlr4与python3运行时一起使用。在我试图解析的语言中,有许多操作(大约50个)接受表单中固定数量的参数 OPNAME [ parameter1, parameter2, parameter3 ]

我以前有一个语法,有这样的规则:

statement: OP1 '[' NUM ']'
    | OP2 '[' NUM ',' NUM ']'
    | OP3 '[' NUM ',' NUM ',' NUM ']'
    | OP2or3 (('[' NUM ',' NUM ']')|('[' NUM ',' NUM ',' NUM ']'))
    ;

然而,为了更清楚,我决定制定一个子规则 parameter[n] 完全接受 n 参数。因此,我的(完整示例)语法如下:

grammar test;

program: (statement? NEWLINE)* EOF;

statement: OP1 parameter[1]
    | OP2 parameter[2]
    | OP3 parameter[3]
    | OP2or3 (parameter[2]|parameter[3])
    ;

parameter[n]
locals[i = 1]
    : '[' NUM 
        ( ',' NUM {$i += 1} )*
      ']' 
      {$i == $n}?
    ;



OP1     : 'OP1'     ;
OP2     : 'OP2'     ;
OP3     : 'OP3'     ;
OP2or3  : 'OP2or3'  ;

NUM     : ('0'..'9')+;
NEWLINE : '\n'      ;
WS      : [ \t\r] -> channel(1);

在以下对象上运行此语法 testfile.txt 几乎可以。我在OP1、OP2和OP3中测试了或多或少的参数,如果我没有确切对应的参数数量,那就失败了。 然而,这对OP2或OP3不起作用,OP2或3在3个参数下总是失败 。我猜antlr解析器首先尝试使用2个参数进行检查,失败谓词,然后无法正确回溯(错误消息为 Error at [5:16] : rule parameter failed predicate: {$i == $n}? ). 的内容 测试文件.txt :

OP1 [1] 
OP2 [32, 52]
OP3 [1, 2, 3]
OP2or3 [1, 2]
OP2or3 [1, 2, 3]

我试图用一个更明确的规则替换入口处的谓词,但这仍然不起作用(错误消息是 Error at [5:7] : no viable alternative at input '[' )

parameter[n]
    : {$n == 1}? '[' NUM ']'
    | {$n == 2}? '[' NUM ',' NUM ']'
    | {$n == 3}? '[' NUM ',' NUM ',' NUM ']'
    ;

作为参考,下面是我用来测试语法的python代码:

import codecs
from antlr4 import *
from antlr4.error.ErrorListener import ErrorListener
from testParser import testParser as Parser
from testLexer import testLexer as Lexer

class SimpleErrorThrower(ErrorListener):
    def syntaxError(self, recognizer, offendingSymbol, line, column, msg, e):
        msg = msg.replace('\n', '\\n')
        raise RuntimeError("Error at [%s:%s] : %s" % (line, column, msg))

def load_code(filename):
    return codecs.decode(open(filename, 'rb').read(), 'utf-8')

def ParseFromRule(input_string, rule_to_call='program'):
    '''Try to parse a given string (case insensitive) from a given rule.
        Raises 'AttrivuteError' if rule does not exist.
        Raises 'ParsingException' if parsing failed.
        Returns the parse tree if parsing was successfull.'''
    source = InputStream(input_string)
    lexer = Lexer(source)
    stream = CommonTokenStream(lexer)
    parser = Parser(stream)
    parser.removeErrorListeners()
    parser.addErrorListener(SimpleErrorThrower())
    parseTree = getattr(parser, rule_to_call)()
    return parseTree


if __name__ == '__main__':
    from argparse import ArgumentParser

    args = ArgumentParser()
    args.add_argument("-p", "--print", help="Print resulting tree.", action='store_true')
    args.add_argument("filename", metavar="Source filename", help="file containing the code to test.", type=str)
    options = args.parse_args()

    input_string = load_code(options.filename)
    try:
        tree = ParseFromRule(input_string, 'program')
    except RuntimeError as e:
        print(str(e))
        exit(1)

    if options.print:
        print(tree.toStringTree(recog=tree.parser))

这是我的 Makefile :

ANTLR_CP=/usr/local/bin/antlr-4.5.1-complete.jar
ANTLR=java -Xmx500M -cp "$(ANTLR_CP):$$CLASSPATH" org.antlr.v4.Tool

all: testParser.py

clean:
    rm -f *Lexer.py *Listener.py *Parser.py *.tokens *.pyc

testParser.py: *.g4
    $(ANTLR) -Dlanguage=Python3 test.g4

你知道我能不能制定一个规则吗 参数[n] 这也适用于 OP2or3 ? 有了这个子规则确实有助于提高清晰度,因为规则往往会经常更改(每隔几个月就会添加或删除一些运算符)