英文:
Antlr3: building parse tree for qualified names
问题
我无法找到一个与我的问题相近的问题/回答，能够帮助我解决我的问题。因此，我在这里发布这个问题。
我正在尝试为限定名称（qualified names）构建一个解析树。下面的示例显示了一个例子。
例如，
1. foo_boo.aaa.ccc1_c [![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/O5LAx.png
在这里，我有用点号分隔的单词。我正在使用antlr3，以下是我的语法。
parse
: expr
;
list_expr : <我移除了这里的语法>
SimpleType : ('a'..'z'|'A'..'Z'|'')('a'..'z'|'A'..'Z'|'0'..'9'|'')*
;
QualifiedType : SimpleType | SimpleType ('\.' SimpleType)+;
expr : list_expr
| QualifiedType
| union_expr;
/*------------------------------------------------------------------
- 词法分析器规则
------------------------------------------------------------------/
WHITESPACE : ( '\t' | ' ' | '\r' | '\n'| '\u000C' )+ { $channel = HIDDEN; } ;
在这里，`SimpleType` 表示一个单词的语法。我的要求是为 `QualifiedType` 构建语法。上面给出的当前语法不能如预期般工作（`QualifiedType : SimpleType | SimpleType ('\.'SimpleType)+;`）。**如何编写正确的限定名称（由点号分隔的单词）的语法？**
英文:
I couldn't find a question/answer that comes close to helping with my issue. Therefore, I am posting this question here.
I am trying to build a parse tree for qualified names. The below example shows an example.
E.g.,
- foo_boo.aaa.ccc1_c
Here I have dot separated words. I am using antlr3 and below is my grammer.
parse
: expr
;
list_expr : <I removed the grammar here>
SimpleType : ('a'..'z'|'A'..'Z'|'_')('a'..'z'|'A'..'Z'|'0'..'9'|'_')*
;
QualifiedType : SimpleType | SimpleType ('\.' SimpleType)+;
expr : list_expr
| QualifiedType
| union_expr;
/*------------------------------------------------------------------
* LEXER RULES
*------------------------------------------------------------------*/
WHITESPACE : ( '\t' | ' ' | '\r' | '\n'| '\u000C' )+ { $channel = HIDDEN; } ;
Here, SympleType represents grammar for a word. My requirement is to build the grammar for the QualifiedType. The current grammar given in above is not working as expected (QualifiedType : SimpleType | SimpleType ('\.'SimpleType)+;). How to write correct grammar for Qualified names (Dot separated words)?
答案1
得分: 0
将 QualifiedType 修改为解析规则而不是词法规则：
qualifiedType : SimpleType ('.' SimpleType)*;
同时，'\'.' 不需要转义：'.' 即可。
编辑
您需要将输出设置为 AST 并应用一些树重写规则，以使其正常工作。以下是一个快速示例：
grammar T;
options {
output=AST;
}
tokens {
Root;
QualifiedName;
}
parse
: qualifiedType EOF -> ^(Root qualifiedType)
;
qualifiedType
: SimpleType ('.' SimpleType)* -> ^(QualifiedName SimpleType+)
;
SimpleType
: ('a'..'z' | 'A'..'Z' | '_') ('a'..'z' | 'A'..'Z' | '0'..'9' | '_')*
;
如果您现在运行以下代码：
import org.antlr.runtime.*;
import org.antlr.runtime.tree.CommonTree;
import org.antlr.runtime.tree.DOTTreeGenerator;
import org.antlr.stringtemplate.StringTemplate;
public class Main {
public static void main(String[] args) throws Exception {
TLexer lexer = new TLexer(new ANTLRStringStream("foo_boo.aaa.ccc1_c"));
TParser parser = new TParser(new CommonTokenStream(lexer));
CommonTree tree = (CommonTree)parser.parse().getTree();
DOTTreeGenerator gen = new DOTTreeGenerator();
StringTemplate st = gen.toDOT(tree);
System.out.println(st);
}
}
您将获得一些 DOT 输出，该输出对应以下 AST：
英文:
Make QualifiedType a parser rule instead of a lexer rule:
qualifiedType : SimpleType ('.' SimpleType)*;
Also, '\.' does not need an escape: '.' is OK.
EDIT
You'll have to set the output to AST and apply some tree rewrite rules to make it work properly. Here's a quick demo:
grammar T;
options {
output=AST;
}
tokens {
Root;
QualifiedName;
}
parse
: qualifiedType EOF -> ^(Root qualifiedType)
;
qualifiedType
: SimpleType ('.' SimpleType)* -> ^(QualifiedName SimpleType+)
;
SimpleType
: ('a'..'z' | 'A'..'Z' | '_') ('a'..'z' | 'A'..'Z' | '0'..'9' | '_')*
;
And if you now run the code:
import org.antlr.runtime.*;
import org.antlr.runtime.tree.CommonTree;
import org.antlr.runtime.tree.DOTTreeGenerator;
import org.antlr.stringtemplate.StringTemplate;
public class Main {
public static void main(String[] args) throws Exception {
TLexer lexer = new TLexer(new ANTLRStringStream("foo_boo.aaa.ccc1_c"));
TParser parser = new TParser(new CommonTokenStream(lexer));
CommonTree tree = (CommonTree)parser.parse().getTree();
DOTTreeGenerator gen = new DOTTreeGenerator();
StringTemplate st = gen.toDOT(tree);
System.out.println(st);
}
}
you'll get some DOT output, which corresponds to the following AST:
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