在嵌套函数模板中使用模板类型名

Question

Alright, I'm struggling with templates. In this question I learned, that it is not possible to pass a type specifier to a function at all, so my next approach is passing the type inside <> .

Imagine a function template foo<U>() , which is member function of a template class A . So if I create an Object A<T> a I can call a.foo<U>() with any type.

How do I have to write an equivalent function template so I can pass a and U like wrappedFoo<U>(a) ?

IMPORTANT Needs to be C++98 compliant

Answer 1

You might do the following:

template <typename U, typename T>
XXXX /* See below */
WrappedFoo(/*const*/ A<T>& a)
{
    return a.template foo<U>();
}

The hard part is the return type without decltype of C++11.

So if the return type really depends of parameters type, you can create a trait, something like:

template <typename U, typename T>
struct Ret
{
    typedef U type;
};

template <typename T> struct Ret<T, A<T> >
struct Ret
{
    typedef bool type;
};

And then replace XXXX by typename Ret<U, T>::type