函数模板专用化语法聚合模板化类型

Question

关于函数模板专用化，我需要一些语法方面的帮助。 下面是一个简化的方案：

基本标头：

template <typename T>
void Foo(T* t) { TRACE("Default Foo impl"); }  // <-- default implementation
template <typename T>
struct Base
{
    explicit Base()
    {
        static_assert(std::is_base_of<Base, T>::value, "T must derive from Base");
        Foo(static_cast<T*>(this));
    }
};

Derived_X标头：

struct Derived_X : public Base<Derived_X>
{
    explicit Derived_X() : Base<Derived_X> { } { }
};
// no specialization will be implemented --> using default

Derived_Y标头：

struct Derived_Y : public Base<Derived_Y>
{
    explicit Derived_Y() : Base<Derived_Y> { } { }
};
template <>  // Foo specialization for Derived_Y
void Foo<Derived_Y>(Derived_Y* t)
{
    Foo(static_cast<Base<Derived_Y>*>(t));  // <-- call default impl
    TRACE("Derived_Y Foo impl");
}

Derived_Z标头：

template <typename T>
struct Derived_Z : public Base<T>
{
    explicit Derived_Z() : Base<T> { }
    {
        static_assert(std::is_base_of<Derived_Z, T>::value, "T must derive from Derived_Z");
    }
};
/*  What does this specialization look like?
template <typename T>
void Foo<Derived_Z<T>>(Derived_Z<T>* t)
{
    Foo(static_cast<Base<T>*>(t));  // <-- call default impl
    TRACE("Derived_Z<T> Foo impl");
}
// */

最派生标头：

struct MostDerived : public Derived_Z<MostDerived>
{
    explicit MostDerived() : Derived_Z<MostDerived> { } { }
};
template <>
void Foo<MostDerived>(MostDerived* t)
{
    Foo(static_cast<Derived_Z<MostDerived>*>(t));  // <-- call Derived_Z impl
    TRACE("MostDerived Foo impl");
}

用法：

int main()
{
    Derived_X dx { };   // <-- "Default Foo impl"
    Derived_Y dy { };   // <-- "Default Foo impl" && "Derived_Y Foo impl"
    MostDerived md { }; // <-- "Default Foo impl" && "MostDerived Foo impl"
}

我无法确定如何为Derived_Z专门Foo。 任何帮助将不胜感激！

Answer 1

通常认为专用函数模板的形式很差。您将无法部分专业化(正如您所注意到的(，并且它们不会在过载分辨率中被考虑。相反，只需创建重载，让重载解决处理其余的工作。

// no template at all for Derived_Y
void Foo(Derived_Y* t)
{
    Foo(static_cast<Base<Derived_Y>*>(t));  // <-- call default impl
    TRACE("Derived_Y Foo impl");
}
// a regular template (no specialization) for Derived_Z<T>
template <typename T>
void Foo(Derived_Z<T>* t)
{
    Foo(static_cast<Base<T>*>(t));  // <-- call default impl
    TRACE("Derived_Z<T> Foo impl");
}
// again, no template for MostDerived
void Foo(MostDerived* t)
{
    Foo(static_cast<Derived_Z<MostDerived>*>(t));  // <-- call Derived_Z impl
    TRACE("MostDerived Foo impl");
}

现在，您可能需要考虑将基本实现更改为仅接受Base<T>*而不是T*。假设您有派生自 Derived_Y 的Derived_Y2，但您尚未定义 Foo(Derived_Y2*) 的重载。使用指向Derived_Y2的指针调用Foo()将进入Foo(T*)，T被推导为Derived_Y2，因为这比Foo(Derived_Y*)更好的匹配

struct Derived_Y2 : Derived_T { };
Derived_Y2 y2; // "Default Foo impl"

通过将基本实现更改为：

template<class T>
void Foo(Base<T>*) { TRACE("Default Foo impl"); }

当给定指向Derived_Y2的指针时，Foo(Derived_Y*)现在将是一个更好的匹配，因为它正如他们所说更专业。

Answer 2

我无法确定如何为Derived_Z专门化Foo。

这是因为您无法部分专用化模板函数。

但是您可以部分专化class/struct.

所以我建议使用一个助手struct

template <typename T>
struct bar
 {
   static void func (T *)
    { std::cout << "default bar" << std::endl; }
 };
template <>
struct bar<Derived_Y>
 {
   static void func (Derived_Y *)
    { std::cout << "Derived_Y bar" << std::endl; }
 };
template <typename T>
struct bar<Derived_Z<T>>
 {
   static void func (Derived_Z<T> *)
    { std::cout << "Derived_Z<T> bar" << std::endl; }
 };

Foo()只是成为

template <typename T>
void Foo (T * t)
 { bar<T>::func(t); }