[xsd-users] xsdcxx-musicxml

[Mario Lang]

Boris Kolpackov <boris at codesynthesis.com> writes:

> Mario Lang <mlang at delysid.org> writes:
>
>> <xs:sequence>
>>   <xs:choice>
>>     <xs:sequence>
>>       <xs:element name="a".../>
>>       <xs:element name="c".../>
>>       <xs:element name="d".../>
>>     </xs:sequence>
>>     <xs:sequence>
>>       <xs:element name="b".../>
>>       <xs:element name="c".../>
>>       <xs:element name="e".../>
>>     </xs:sequence>
>>     <xs:sequence>
>>       <xs:element name="c".../>
>>       <xs:element name="d".../>
>>       <xs:element name="e".../>
>>     </xs:sequence>
>>   </xs:choice>
>>   <xs:element.../>...
>> </xs:sequence>
>
> To serialize, e.g., 'c' in the correct order, XSD would need to know
> which arm of the choice it is.
> Since XSD "flattens" choices, this is not possible in the current
> architecture. In this particular case you can kind of infer which arm
> it is based on which elements are present/absent. But that would be
> impossible to extend to the general case.

OK, thnaks, I trust your analysis on this one.
Just wanted to make sure we are not accidentally misunderatnding
eachotehr.

>> I have a feeling that all that is missing, is to fix the "sorting" order
>> of the flattened structure.  Or, maybe, to detect identical elements in
>> a choice of sequences.  Because, in the <choice> above, <c> is *always*
>> required to be present, <a> and <b> are mutually exclusive, and <d> and
>> <e> are only allowed if one of <a> or <b> is present.  Again, to me, it
>> looks like the serialization order is rather unambigious from the
>> grammar given.
>
> I would suggest that you customize the serializer operator for this
> type. Implementing this in XSD will be non-trivial, to put it mildly.

OK.  A customization works as well for me.

http://github.com/mlang/xsdcxx-musicxml/

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