28901 2015-07-05 20:52:50 +0000 [XP 3.1] Unary Lookup 2015-07-14 21:53:47 +0000 1 1 1 Unclassified XPath / XQuery / XSLT XPath 3.1 Candidate Recommendation PC All RESOLVED FIXED P2 normal --- 1 mike jonathan.robie public-qt-comments oldest_to_newest 121609 0 mike 2015-07-05 20:52:50 +0000 Section 3.11.3.1 contains the text If the context item is an array, the UnaryLookup operator is equivalent to the following expression: for $k in 1 to array:size(.) return .($k) This is true only for the wildcard form of the operator, but that is covered later in the section. I believe that the previous statement If the context item is a map, the UnaryLookup operator is equivalent to the following expression: for $k in KS return .($k) is true also when the context item is an array. Using nested lists to make clear the scope of the conditionals in this section would make it much clearer. I would be inclined to structure it as follows, even though this involves duplication: 1. If the context item is a map: 1.1 If KeySpecifier is an NCName: 1.2 If KeySpecifier is an IntegerLiteral: 1.3 If KeySpecifier is a ParenthesizedExpr: 1.4 If KeySpecifier is a wildcard ("*") 2. If the context item is an array: 2.1 If KeySpecifier is an NCName: (always an error) 2.2 If KeySpecifier is an IntegerLiteral: 2.3 If KeySpecifier is a ParenthesizedExpr: 2.4 If KeySpecifier is a wildcard ("*"): 3. Otherwise (error) 121610 1 mike 2015-07-05 20:58:36 +0000 In 3.11.13.2 we read The semantics of E?S are equivalent to for $a in E, $b in S return $a($b) , where E is an expression on the left of the postfix ? operator and ?S is the unary lookup operator described in 3.11.3.1 Unary Lookup. This doesn't work. "?S" isn't a unary lookup operator. I think this should be The semantics of E?S are equivalent to for $a in E return $a ! ?S where ?S uses the unary lookup operator described in 3.11.3.1 Unary Lookup 121614 2 mike 2015-07-06 07:15:13 +0000 Note also, I believe that if $A is an array, then $A?* has the same result as array:flatten($A); it would be useful to point out this relationship, or even to use it as the normative specification of $A?* 122042 3 jonathan.robie 2015-07-14 21:53:47 +0000 (In reply to Michael Kay from comment #0) > Using nested lists to make clear the scope of the conditionals in this > section would make it much clearer. I would be inclined to structure it as > follows, even though this involves duplication: > > 1. If the context item is a map: > 1.1 If KeySpecifier is an NCName: > 1.2 If KeySpecifier is an IntegerLiteral: > 1.3 If KeySpecifier is a ParenthesizedExpr: > 1.4 If KeySpecifier is a wildcard ("*") > 2. If the context item is an array: > 2.1 If KeySpecifier is an NCName: (always an error) > 2.2 If KeySpecifier is an IntegerLiteral: > 2.3 If KeySpecifier is a ParenthesizedExpr: > 2.4 If KeySpecifier is a wildcard ("*"): > 3. Otherwise (error) That's the approach I took.