具有依赖参数类型的概念约束成员函数

我用概念做了一些实验,我试图得到约束成员函数,只有在满足概念的情况下才能实例化这些函数:

template <typename T>
concept Fooable = requires(T o)
{
    o.foo(uint());
};

template <typename T>
concept Barable = requires(T o)
{
    o.bar(uint());
};

class Foo
{
public:
    using FooType = int;
    void foo(uint) {}
};

class Bar
{
public:
    using BarType = double;
    void bar(uint) {}
};

template <typename T>
class C
{
public:
    void fun(typename T::FooType t) requires Fooable<T> {}
    void fun(typename T::BarType t) requires Barable<T> {}
};

int main()
{
    C<Foo> f;
}

main.cpp: error: no type named 'BarType' in 'Foo'
main.cpp: error: no type named 'BarType' in 'Foo'
        void fun(typename T::BarType t) requires Barable<T> {}
                 ~~~~~~~~~~~~^~~~~~~
main.cpp: note: in instantiation of template class 'C<Foo>' requested here
        C<Foo> f;
               ^

然而,由于我宣布 C 具有的对象 Foo fun 具有 BarType 不会实例化。