I am making a template class with an inner utility class. All specializations of the template want the same inner class:
template<...> class Outer {
class Inner { };
};
That gives me Outer<...>::Inner
but I want all Inner
to be the same type, as if I'd just written:
class Inner { };
template <...> class Outer { };
or if Outer
were simply not a template class:
class Outer {
class Inner { };
};
giving me Outer::Inner
. I'd like to have Outer::Inner
wor开发者_如何学Gok for all Outer<>
if that's possible (just for namespace/clarity reasons). Otherwise of course I can just move Inner
out.
The nested class can be a non-template, but every instantiation of the template will have its own nested class because they're (otherwise) unrelated types. You can do
namespace detail {
class Inner {};
} // detail
template<...>
class Outer {
typedef detail::Inner Inner;
};
The way I've done this in the past is using inheritance:
class DummyBase{
protected:
class Inner{
//etc...
};
};
template<...> class Outer : public DummyBase{
//etc...
};
It will be unique for each instantiation of Outer. I.e.,
Outer<int>::Inner will be a different type from Outer<double>::Inner
精彩评论