使用antlr3将堆叠的比较表达式解析为逻辑合取树 [英] Parse stacked comparison expression into logical Conjunction Tree with antlr3

问题描述

当我尝试解析堆积的算术比较表达式时，我遇到了一个问题:

I have run into a problem, when i tried to parse a stacked arithmetic comparison expression:

"1<2<3<4<5" 

进入逻辑连接树:

CONJUNCTION(COMPARISON(1,2,<) COMPARISON(2,3,<) COMPARISON(3,4,<) COMPARISON(4,5,<))

Antlr3 Tree Rewrite规则中是否有一种方法可以迭代匹配的标记并以目标语言从它们创建结果树(我使用的是Java)?因此，我可以根据匹配的加法"标记的元素x，x-1创建比较节点.我知道我可以引用规则的最后结果，但是那样一来，我只会得到嵌套的比较规则，这不是我想要的.

Is there a way in Antlr3 Tree Rewrite rules to iterate through matched tokens and create the result Tree from them in the target language (I'm using java)? So i could make COMPARISON nodes from element x, x-1 of matched 'addition' tokens. I know i can reference the last result of a rule but that way i'd only get nested COMPARISON rules, that's not what i wish for.

/This is how i approached the problem, sadly it doesn't do what i would like to do yet of course.

fragment COMPARISON:;

operator
:
('<'|'>'|'<='|'>='|'=='|'!=')
;

comparison
@init{boolean secondpart = false;}
:
e=addition (operator {secondpart=true;} k=addition)* 
-> {secondpart}? ^(COMPARISON ^(VALUES addition*) ^(OPERATORS operator*))
-> $e
;

//Right now what this does is:
tree=(COMPARISON (VALUES (INTEGERVALUE (VALUE 1)) (INTEGERVALUE (VALUE 2)) (INTEGERVALUE (VALUE 3)) (INTEGERVALUE (VALUE 4)) (INTEGERVALUE (VALUE 5))) (OPERATORS < < < <))



//The label for the CONJUNCTION TreeNode that i would like to use:
fragment CONJUNCTION:;  

推荐答案

我通过编写实际的树构建Java代码，想出了一个解决此问题的讨厌的方法:

I came up with a nasty solution to this problem by writing actual tree building java code:

grammar testgrammarforcomparison;

options {
  language = Java;
  output = AST;
}

tokens
{
 CONJUNCTION;
 COMPARISON;
 OPERATOR;
 ADDITION;
}


WS
:
    ('\t' | '\f' | ' ' | '\r' | '\n' )+
    {$channel = HIDDEN;}
;

comparison
@init 
{ 
  List<Object> additions = new ArrayList<Object>();
  List<Object> operators = new ArrayList<Object>();
  boolean secondpart = false;
}
:
(( e=addition {additions.add(e.getTree());} ) ( op=operator k=addition {additions.add(k.getTree()); operators.add(op.getTree()); secondpart = true;} )*) 
{
  if(secondpart)
  {
      root_0 = (Object)adaptor.nil();

            Object root_1 = (Object)adaptor.nil();
            root_1 = (Object)adaptor.becomeRoot(
            (Object)adaptor.create(CONJUNCTION, "CONJUNCTION")
            , root_1); 

      Object lastaddition = additions.get(0);

            for(int i=1;i<additions.size();i++)
      {
        Object root_2 = (Object)adaptor.nil();
        root_2 = (Object)adaptor.becomeRoot(
        (Object)adaptor.create(COMPARISON, "COMPARISON")
        , root_2); 

        adaptor.addChild(root_2, additions.get(i-1)); 
        adaptor.addChild(root_2, operators.get(i-1));
        adaptor.addChild(root_2, additions.get(i)); 

        adaptor.addChild(root_1, root_2);
      }
  adaptor.addChild(root_0, root_1);
  }
  else
  {
   root_0 = (Object)adaptor.nil();
   adaptor.addChild(root_0, e.getTree());
  }
}
;
/** lowercase letters */
fragment LOWCHAR
:   'a'..'z';
/** uppercase letters */
fragment HIGHCHAR
:   'A'..'Z';
/** numbers */
fragment DIGIT
:   '0'..'9';

fragment LETTER
: LOWCHAR
| HIGHCHAR
;

IDENTIFIER
:
 LETTER (LETTER|DIGIT)*
;

addition
:
 IDENTIFIER ->^(ADDITION IDENTIFIER)
;

operator 
:
 ('<'|'>') ->^(OPERATOR '<'* '>'*)
;

parse
:
 comparison EOF
;

用于输入

"DATA1 < DATA2 > DATA3" 

这将输出树，例如:

如果你们知道任何更好的解决方案，请告诉我有关他们的信息

If you guys know any better solutions, please tell me about them

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