30317 2018-11-08 20:52:59 +0000 [XP31] Subtype rules for maps tested against functions conflict, because of zero-or-one cardinality in rule 35 2018-11-08 20:52:59 +0000 1 1 1 Unclassified XPath / XQuery / XSLT XPath 3.1 Recommendation PC Windows NT NEW P2 normal --- 1 abel.online jonathan.robie public-qt-comments oldest_to_newest 129383 0 abel.online 2018-11-08 20:52:59 +0000 A map item can have a value of type item()*. The rules for the subtype relationship (2.5.6.2 The judgement subtype-itemtype(Ai, Bi)) for maps state: 35. Ai is map(K, V), and Bi is function(xs:anyAtomicType) as V?. I see two issues with this statement: 1) Bi can be any subtype of function(xs:anyAtomicType) as V?. It doesn't have to be exactly that type. 2) It should be V*, not V? Example: map{'a': (1, 2)} instance of function(xs:int) as item()? a) Current rules The map type of map(K, V) is: map(xs:string, xs:integer*) The corresponding function then becomes: function(xs:anyAtomicType) as xs:integer? Result of the type test would then be true, but that is clearly wrong, considering the co-variance of the return type. b) If we replace V? with V* Now the rewrite is instead: function(xs:anyAtomicType) as xs:integer* Now the instance-of expressions would correctly return true. Saxon gets it right with the (b) scenario, even though that is at odds with the spec. I think (b) is the correct approach. The same error can be found in section 2.5.5.8 Map Test. I propose the rule be rewritten in an erratum, something along these lines: <proposal> 35. Ai is map(K, V), and Bi is function(xs:anyAtomicType) as V* or a subtype thereof. </proposal> This change would, however, lead to map{'a': 1} being no longer a subtype of function(xs:int) as item()? To fix that, we could write it as follows: <proposal> 35. Ai is map(K, VA), and Bi is function(KB) as VB, where subtype(VA, VB) and subtype-itemtype(KB, xs:anyAtomicType). </proposal> This latter syntax would be more in line with rule 28, also on maps and emphasizes the co-variance/contra-variance rules that are in place here.