--- Comment #11 from Stewart Gordon <> 2011-07-08 16:42:24 PDT ---
I'm not sure you get the point.  It's to give the class designer the choice
between in-place modification and creating a new object as the implementation
of op=.

OK, so a op= b ought to be automatically equivalent to a = a op b in the
absence of an opOpAssign.  So the rule would be to define opOpAssign iff you
want it to modify in-place.  But this still doesn't cover cases where whether
the modification is in-place isn't a compile-time constant.  Look at how ~=
works for instance.

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