我正在使用Antlr4与python3运行时 . 在我尝试解析的语言中,有许多操作(大约50个)接受 OPNAME [ parameter1, parameter2, parameter3 ]
形式的固定数量的参数
我曾经有一个像这样的规则的语法:
statement: OP1 '[' NUM ']'
| OP2 '[' NUM ',' NUM ']'
| OP3 '[' NUM ',' NUM ',' NUM ']'
| OP2or3 (('[' NUM ',' NUM ']')|('[' NUM ',' NUM ',' NUM ']'))
;
但是,为了更加清晰,我决定制作一个完全接受 n
参数的子规则 parameter[n]
. 因此,我的(完整示例)语法看起来像这样:
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 == $n}?
;
OP1 : 'OP1' ;
OP2 : 'OP2' ;
OP3 : 'OP3' ;
OP2or3 : 'OP2or3' ;
NUM : ('0'..'9')+;
NEWLINE : '\n' ;
WS : [ \t\r] -> channel(1);
在以下 testfile.txt
上运行此语法几乎可以正常工作 . 我测试过在OP1,OP2和OP3中有更多或更少的参数,如果我没有完全相应的参数数量,那就失败了 . However, that does not work for OP2or3, which always fails for 3 parameters . 我想antlr解析器首先尝试使用2个参数进行检查,谓词失败,然后无法正确回溯(错误消息是 Error at [5:16] : rule parameter failed predicate: {$i == $n}?
) . 内容 testfile.txt
:
OP1 [1]
OP2 [32, 52]
OP3 [1, 2, 3]
OP2or3 [1, 2]
OP2or3 [1, 2, 3]
我尝试用入口处的谓词更明确的规则替换,但仍然不起作用(错误消息是 Error at [5:7] : no viable alternative at input '['
)
parameter[n]
: {$n == 1}? '[' NUM ']'
| {$n == 2}? '[' NUM ',' NUM ']'
| {$n == 3}? '[' NUM ',' NUM ',' NUM ']'
;
有关信息,这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
Do you have any idea if I could make a rule parameter[n] that will also work for OP2or3 ? 让这个子规则确实有助于清晰度,这个规则往往会经常发生变化(每隔几个月就会增加或删除一些运算符)
1 回答
好吧,很抱歉打扰了任何看我问题的人,但我自己找到了答案,使用了python magic . 也许有一天可以帮助某人 . 我重写
parameter[n]
以输入int
或tuple
:请注意语义谓词中的括号 . You need to put thoses parentheses 因为它将在python解析器中被翻译为
not (i == n or i in n)
,并且没有不能正确否定的括号(我想这可能被视为一个antlr4错误 . )因此,现在我的陈述规则是:
并在我的测试文件上工作: