问题描述
我想创建一个没有侦听器和访问者的非常简单的ANTLR4解析器(在Python中),该解析器以任意顺序将书的章节和段落作为输入,并返回high_level
(章节)和{{ 1}}(段落),例如如果我输入low_level
或2 a
,则应打印“ 第2章,a段”。
这是我的a 2
Example.g4
我在终端中执行此操作
grammar Example;
text
: paragraph ;
paragraph
: high_level (WS low_level)?
| low_level WS high_level
;
low_level
: 'a' | 'b' | 'c' ;
high_level
: '1' | '2' | '3';
WS : [ \t\r\n]+ ;
java -jar ~/antlr-4.8-complete.jar -Dlanguage=python3 -no-listener -no-visitor Example.g4
但是,如果我随后运行它并输入from antlr4 import *
from ExampleLexer import ExampleLexer
from ExampleParser import ExampleParser
def main():
while True:
text = InputStream(input(">"))
lexer = ExampleLexer(text)
stream = CommonTokenStream(lexer)
parser = ExampleParser(stream)
tree = parser.text()
query = tree.paragraph()
low_level = query.low_level()
high_level = query.high_level()
print(f"chapter {high_level},paragraph {low_level}")
if __name__ == '__main__':
main()
,我会得到
2 a
有人可以解释我在做什么错吗?我不明白方括号中的数字。
解决方法
这只是RuleContext
显示的一些调试信息(生成的Low_levelContext
和High_levelContext
类从中扩展)。在您的情况下,显示规则的invokingState
和parentCtx
。
看看来源:
class RuleContext(RuleNode):
...
def __str__(self):
return self.toString(None,None)
...
def toString(self,ruleNames:list,stop:RuleContext)->str:
with StringIO() as buf:
p = self
buf.write("[")
while p is not None and p is not stop:
if ruleNames is None:
if not p.isEmpty():
buf.write(str(p.invokingState))
else:
ri = p.getRuleIndex()
ruleName = ruleNames[ri] if ri >= 0 and ri < len(ruleNames) else str(ri)
buf.write(ruleName)
if p.parentCtx is not None and (ruleNames is not None or not p.parentCtx.isEmpty()):
buf.write(" ")
p = p.parentCtx
buf.write("]")
return buf.getvalue()
...
https://github.com/antlr/antlr4/blob/master/runtime/Python3/src/antlr4/RuleContext.py
您没有解释要显示的内容,但是我想规则匹配的文本,在这种情况下,您可以执行以下操作:
print(f"chapter {high_level.getText()},paragraph {low_level.getText()}")