surprising behaviour encountered while covering more cases
...obviously the return type of ExpandFunctor::operator()
was inferred as value, even while the invoked functor, from which
this type was deduced, clearly returns a reference.
Solution is simple not to rely on inference, moreover since we know
the exact type in the enclosing scope, thanks to the refactoring which
made this ExpandFunctor a nested class
NOTE:
as it turned out, this is not a compiler bug,
but works as defined by the language:
on return type inference, the detected type is decayed,
which usually helps to prevent returning a reference to a temporary
Ichthyostega
77c3226948