ANTLR - 带空格的标识符 [英] ANTLR - identifier with whitespace
问题描述
我想要可以包含空格的标识符.
i want identifiers that can contain whitespace.
grammar WhitespaceInSymbols;
premise : ( options {greedy=false;} : 'IF' ) id=ID{
System.out.println($id.text);
};
ID : ('a'..'z'|'A'..'Z')+ (' '('a'..'z'|'A'..'Z')+)*
;
WS : ' '+ {skip();}
;
当我使用IF 语句分析"进行测试时,我得到一个 MissingTokenException 和输出IF 语句分析".
我想,通过使用 greedy=false 我可以告诉 ANTLR 在 'IF' 之后退出并将其作为令牌.但 IF 是 ID 的一部分.有没有办法实现我的目标?我已经尝试了 greed=false 选项的一些变体,但没有成功.
When i test this with "IF statement analyzed" i get a MissingTokenException and the output "IF statement analyzed".
I thought, that by using greedy=false i could tell ANTLR to exit afer 'IF' and take it as a token. But instead the IF is part of the ID.
Is there a way to achieve my goal? I already tried some variations of the greed=false-option, but without success.
推荐答案
我想,通过使用 greedy=false 我可以告诉 ANTLR 在 'IF' 之后退出并将其作为令牌.
I thought, that by using greedy=false i could tell ANTLR to exit afer 'IF' and take it as a token.
不,解析器对令牌的创建没有什么可说的:首先对输入进行令牌化,然后将解析器规则应用于这些令牌.所以设置 greedy=false
没有效果.
No, the parser has nothing to say about the creation of tokens: the input is first tokenized and then the parser rules are applied on these tokens. So setting greedy=false
has no effect.
您可以这样做(创建带有空格的 ID
标记),但这将是一个糟糕的解决方案,其中包含许多谓词,并且词法分析器中的一些自定义方法正在执行手动预测:你真的,真的不想要这个!更简洁的解决方案是在解析器中引入 id
规则,让它匹配一个或多个 ID
标记.
You can do this (creating ID
tokens with white spaces), but it will be a horrible solution with many predicates, and a few custom methods in the lexer doing manual look-aheads: you really, really don't want this! A much cleaner solution would be to introduce a id
rule in your parser and let it match one or more ID
tokens.
grammar WhitespaceInSymbols;
premise
: IF id THEN EOF
;
id
: ID+
;
IF
: 'IF'
;
THEN
: 'THEN'
;
ID
: ('a'..'z' | 'A'..'Z')+
;
WS
: ' '+ {skip();}
;
会将输入的 IF 语句分析 THEN
解析为以下树:
would parse the input IF statement analyzed THEN
into the following tree:
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