In the example code there is "using Length = long;", but long is not portable, shouldn't be this be int32_t?

Replace use of C++ long in example code with int32_t

IDL longs are mapped to C++ int32_t, the example in clause 7.2.4.6 makes the mistake of mapping a typedef long to a using of long instead of int32_t.

Spec says:

If the union has a default case, the default constructor shall initialize the discriminator, and the selected
member field following the initialization rules described in Clause 7.2.4.3.1. If it does not, the default
constructor shall initialize the union to the first discriminant value specified in the IDL definition.

But when there is an implicit default member, that should be selected, for example, the example below (

  enum DataType
  {
    dtEmpty,
    dtLong,
    dtShort,
    dtString,
    dtPoint,
    dtTrack,
    dtGlobal
  };

  union Data switch (DataType)
    {
      case dtLong: long longData;
      case dtShort: short shortData;
      case dtString: string stringData;
      case dtPoint: string pointData;
      case dtTrack: string trackData;
      case dtGlobal: string globalData;
      // by default (implicit), empty union
    };

Defer the resolution of this issue to the resolution of the related IDL RTF issue

This issue is deferred until the resolution IDL43-94.