2720 2006-01-16 21:46:45 +0000 [FS] Wrong / operator precedence in PathExpr normalization rule 2006-10-19 22:57:14 +0000 1 1 1 Unclassified XPath / XQuery / XSLT Formal Semantics 1.0 Last Call drafts All All CLOSED FIXED P2 normal --- 1 villard simeon public-qt-comments oldest_to_newest 7853 0 villard 2006-01-16 21:46:47 +0000 The normalization rule for PathExpr is not consistent with the / operator precedence, which is left-to-right. It should be written this way: [RelativePathExpr / StepExpr]Expr == fs:apply-ordering-mode ( fs:distinct-doc-order-or-atomic-sequence ( let $fs:sequence as node()* := [RelativePathExpr]Expr return let $fs:last := fn:count($fs:sequence) return for $fs:dot at $fs:position in $fs:sequence return [StepExpr]Expr)) 8025 1 simeon 2006-01-28 17:03:05 +0000 I agree with the comment. The rewriting is clearer according to the precedence rule. I do not believe this changes the actual dynamic or static semantics, but makes it clearer. - Jerome 8093 2 simeon 2006-02-01 20:30:44 +0000 Note that the following rule: [ StepExpr // RelativePathExpr ]Expr == [StepExpr / descendant-or-self::node() / RelativePathExpr]Expr Should also be changed in a consistent way, to: [ RelativePathExpr // StepExpr ]Expr == [ RelativePathExpr / descendant-or-self::node() / StepExpr ]Expr - Jerome 8258 3 simeon 2006-02-14 16:32:53 +0000 The XSLT and XQuery WG have accepted your comments, and the issue will be fixed by adopting the changes listed in comments #1 and #2. Best regards, - Jerome