I want to specialize the type parameter of the following template class, which has a type parameter and a template template parameter:
template <
typename T,
template <typename E> class Foo
> class Bar;
I tried every permutation of adding and/or omitting .template and typename in the last line of each of the following snippets, and none compiles:
1.)
template <
template <typename E> class Foo
> class Bar<int, Foo<typename E>>;
2.)
template <
template <typename E> class Foo
> class Bar<int, Foo.template <typename E>>;
3.)
template <
template <typename E> class Foo
> class Bar<int, Foo<E>>;
4.)
template <
template <typename E> class Foo
class Bar<int, Foo.template <E>>;
Why doesn't any of them work?
Regarding the last line of each applicable snippet:
typename clarify E is a type used by class Foo, or can this syntax only be used within the {} body of Bar's class definition?template clarify Foo is a template and therefore prevent the compiler from parsing Foo < as Foo "is less than", or can this syntax only be used within the {} body of Bar's class definition?How can I get this to work?
Doesn't
typenameclarifyEis a type used by classFoo, or can this syntax only be used within the{}body ofBar's class definition?
typename is only used when you are defining a type within a template definition (class could also be used) or when you are accessing a dependent type (a type that depends on a template parameter).
For more information (even about when template is used) see this thread.
How can I get this to work?
The name of a of a type within template template parameter can't actually be used. It's just there as a formality. You have to add another template parameter to your main template instead:
template <
template<typename> class Foo,
typename E
> class Bar<int, Foo<E>> { ... };
Moreover, if this is a specialization of the template Bar, then Bar needs a primary template to specialize:
template<typename T, typename U>
struct Bar;
template <
template<typename> class Foo,
typename E
> class Bar<int, Foo<E>> { ... };
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