Question

关于功能模板专业化，我想帮助一些语法。以下是简化方案：

基本标题：

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标题：

struct Derived_X : public Base<Derived_X>
{
    explicit Derived_X() : Base<Derived_X> { } { }
};

// no specialization will be implemented --> using default

Derived_Y标题：

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标题：

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");
}
// */

MostDerived Header：

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Foo专门化Derived_Z。任何帮助都将非常感激！

Answer 1

通常认为专门化功能模板的形式很差。您将无法部分专业化（如您所注意到的），并且不会在重载决策中考虑它们。相反，只需创建重载，并让重载解析来处理其余的事情。

// 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Derived_Y2派生自Derived_Y，但您没有为Foo(Derived_Y2*)定义重载。然后使用指向Foo()的指针调用Derived_Y2将转到Foo(T*)，T被推断为Derived_Y2，因为这是比{{更好的匹配1}}

Foo(Derived_Y*)

将基本实现更改为：

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

当给出指向

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

的指针时，

Foo(Derived_Y*)现在将更好地匹配，因为它们正如他们所说的那样更专业。

Answer 2

我无法确定如何为Derived_Z专门化Foo。

这是因为你不能部分专门化模板功能。

但您可以部分专门化class / struct。

所以我建议使用帮助器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; }
 };

和Foo()只是变成

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

功能模板专业化语法聚合模板化类型

2 个答案: