[antlr-interest] Re: Tree transformation
lgcraymer
lgc at mail1.jpl.nasa.gov
Sun Nov 16 12:26:52 PST 2003
Arnar--
Take a look at <http://www.antlr.org/doc/trees.html#Action%
20Translation>. ANTLR labels get transformed during translation; if
I remember correctly, #label translates to labelAST and #label_in to
label. You are probably reusing the input tree by accident here.
Try replacing list with #list in the action.
--Loring
--- In antlr-interest at yahoogroups.com, "Arnar Birgisson"
<arnarb at o...> wrote:
> Hello again..
>
> I solved this problem by rewriting the transformation so that it
worked
> without the !. However, I keep hitting this wall in other places,
and
> now I'm completely stuck. Consider those (simplified) constructs
in my
> language
>
> f -> procedure()
> var x
> body
> expression1,
> block
> x := \incr x,
> \print [1,2,3]
> endblock,
> x := \incr ,
> \print [1,2,3]
> endbody
>
> Now.. the expressions "\incr x" and "[1,2,3]" have such trees:
>
> #([OPERATOR,"incr"] x)
> #([LIST,"["] #([EXPR_LIST] 1 2 3) )
>
> and I have rules in my tree transformer that changes them to the
> equivalent of the expressions "incr(x)" (function call) and
> "mk_pair(1,mk_pair(2,mk_pair(3,[])))". This transformation works
very
> well for the second pair of those expressions.
>
> Now I want to make a transformation for turning
>
> BLOCK
> |
> EXPR_LIST
> |
> expr1 - expr2 - ... - exprN
>
> to
>
> expr1 - expr2 - ... - exprN.
>
> For this I have the rule alternative
>
> |! #(BLOCK list:expr_list)
> {
> ## = list->getFirstChild();
> }
>
> This seems to work except that the transformations inside this
instance
> of expr_list don't get executed. Therefore, the transformation of
the
> code above is applied only to the second pair of afformentioned
> expressions.
>
> I'm using exactly the same rule (expr_list) to traverse the list of
> expressions whether they are inside the procedure body, or inside a
> block. I'm stumped!
>
> If I remove the ! I get two copies of the subtree, one where
> transformations have been applied, and one where they havent...
>
> Am I doing something terribly wrong or is this unexpected
behaviour?
>
> Arnar
>
>
> > -----Original Message-----
> > From: mzukowski at y... [mailto:mzukowski at y...]
> > Sent: 14. nóvember 2003 18:39
> > To: antlr-interest at yahoogroups.com
> > Subject: RE: [antlr-interest] Tree transformation
> >
> >
> > Hard to tell. I recommend using -traceParser and following
> > through the code
> > to see what's happening.
> >
> > Monty
> >
> > -----Original Message-----
> > From: Arnar Birgisson [mailto:arnarb at o...]
> > Sent: Friday, November 14, 2003 7:24 AM
> > To: antlr-interest at yahoogroups.com
> > Subject: [antlr-interest] Tree transformation
> >
> > Hello..
> >
> > I'm having some trouble I can't figure out, possibly because I'm
doing
> > something stupid.
> >
> > I have this rule in a tree parser for transforming loops:
> >
> > loop_stmt
> > : #(L_LOOP stmt_list)
> > | #(L_WHILE expr stmt_list)
> > |! #(L_FOR init:stmt_list test:expr incr:stmt_list
> > body:stmt_list)
> > {
> > /* this changes "for" loops to "while" loops
*/
> > antlr::RefAST newbody;
> > antlr::RefAST lastBodyStmt =
> > body->getFirstChild();
> > if (antlr::nullAST == lastBodyStmt) {
> > newbody = incr;
> > } else {
> > while (antlr::nullAST !=
> > lastBodyStmt->getNextSibling())
> > lastBodyStmt =
> > lastBodyStmt->getNextSibling();
> >
> > lastBodyStmt->setNextSibling(incr->getFirstChild());
> > newbody = body;
> > }
> > antlr::RefAST l = #([L_WHILE,"while"], test,
> > newbody);
> > antlr::RefAST lastInitStmt =
> > init->getFirstChild();
> > if (antlr::nullAST == lastInitStmt) {
> > ## = l;
> > } else {
> > while (antlr::nullAST !=
> > lastInitStmt->getNextSibling())
> > lastInitStmt =
> > lastInitStmt->getNextSibling();
> > lastInitStmt->setNextSibling(l);
> > ## = init->getFirstChild();
> > }
> > }
> > ;
> >
> > Now, stmt_list is a simple rule
> >
> > stmt_list
> > : #(STMT_LIST (stmt)*)
> > ;
> >
> > and the stmt rule is a big rule, with one alternative being this
> > (note that in my language there is no difference between
> > statements and
> > expressions):
> >
> > |! #(OPERATOR s1:expr s2:expr)
> > {
> > /* this changes "x <op> b" to the function
call
> > "<op>(x,y)"
> > #OPERATOR->setType(ID);
> > ## = #([OPEN_PAR,"("], ADGERD,
> > #([stmt_list,"params"], s1, s2));
> > }
> >
> > Now, this alternative successfully transforms operator
statements to
> > function alls when they are top level statements in functions
> > (accessed
> > throught stmt_list), but when they're in a for-loop body
> > (named "body")
> > in the above rule, no transformation takes place, i.e. #
(OPERATOR expr
> > expr) is left as is.
> >
> > I've tried removing the ! in the for-loop rule but that doesn't
help..
> > the transformation doesn't take place.
> >
> > Any ideas?
> >
> > Arnar
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
> > http://docs.yahoo.com/info/terms/
> >
> >
> >
> >
> > Your use of Yahoo!
> > Groups is subject to http://docs.yahoo.com/info/terms/
> >
> >
Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
More information about the antlr-interest
mailing list