mbeckerle commented on a change in pull request #207: Added support for
enumerations and TypeValueCalc
URL: https://github.com/apache/incubator-daffodil/pull/207#discussion_r275127919
##########
File path:
daffodil-core/src/main/scala/org/apache/daffodil/dsom/ChoiceGroup.scala
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@@ -139,6 +167,49 @@ abstract class ChoiceTermBase(
ev
}
+ // If choiceDispatchKeyKind is byType, verify that all our children share a
repType,
+ // and use that. Otherwise, there is no need to associate a repType with
this choice
+ override final lazy val optRepTypeFactory: Option[SimpleTypeDefFactory with
NamedMixin] = defaultableChoiceDispatchKeyKind match {
+ case ChoiceKeyKindType.ByType => {
+ val branchReptypes: Seq[SimpleTypeDefFactory with NamedMixin] =
groupMembers.map(term => {
+ term match{
+ case e: ElementDeclMixin => e.typeDef match{
+ case t: SimpleTypeDefBase => t.optRepTypeDefFactory match{
+ case None => SDE("When <xs:choice> has
choiceBranchKey=\"byType\", all branches must have a type which defines a
repType")
+ case Some(x) => x
+ }
+ case _ : SimpleTypeBase => SDE("When <xs:choice> has
choiceBranchKey=\"byType\", no branch can have a primitive xsd type")
+ case _ => SDE("When <xs:choice> has choiceBranchKey=\"byType\",
all branches must be a simple type")
+ }
+ case _ => SDE("When <xs:choice> has choiceBranchKey=\"byType\",
all branches must be a simple element")
+ }
+ })
+ val ans = branchReptypes.reduce((a, b) => {
+ /*
+ * Since we inline all type definitions, the repTypes will not be
literally equal here, even when they are literally
+ * the same type in schema. Instead of checking for literal equality,
we compare their qNames.
+ */
+ if (a.namedQName != b.namedQName) {
+ SDE("All children of an <xs:choice> with
choiceDispatchKeyKind=byType must have the same reptype")
+ }
+ a
+ })
+ Some(ans)
+ }
+ case ChoiceKeyKindType.Speculative => None
+ case ChoiceKeyKindType.Explicit => None
+ case ChoiceKeyKindType.Implicit => None
+ }
+
+ /*
+ * Logically speaking, a choice group does have a repValueSet, which can be
found as either the union
+ * of that of its choices, or the union of its choiceBranchKey.
+ * In theory, these are both computable, but coersing the choiceBranchKeys
into the appropriate type
+ * is more work than justified for an attribute that we do not actually need.
Review comment:
I didn't understand the comment above.
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