c++ - 函数模板特化语法聚合模板化类型

Question

关于函数模板特化，我需要一些语法方面的帮助。以下是一个简化的场景:

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 header :

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 header :

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 Derived_Z 专门化 Foo。任何帮助将不胜感激!

Answer 1

特化函数模板通常被认为是一种糟糕的形式。您将无法部分特化(正如您所注意到的)，并且在重载解决方案中不会考虑它们。相反，只需创建重载，让重载解析处理其余部分。

// 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，您可能要考虑更改基本实现以仅接受 Base<T>*而不是 T* .假设你有 Derived_Y2源自 Derived_Y , 但您尚未为 Foo(Derived_Y2*) 定义重载.打电话Foo()使用指向 Derived_Y2 的指针然后将转到Foo(T*)与 T被推断为 Derived_Y2 ，因为这是比 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"); }

Foo(Derived_Y*)当给出一个指向 Derived_Y2 的指针时，现在将是一个更好的匹配，因为正如他们所说，它更专业。