我听说c++的类成员函数模板不能是虚的。这是真的吗?

如果它们可以是虚拟的,那么有什么场景可以使用这样的函数呢?


当前回答

如果预先知道模板方法的类型集,则'虚拟模板方法'有一个变通方法。

为了说明这个想法,在下面的例子中只使用了两种类型(int和double)。

在那里,一个“虚拟”模板方法(Base:: method)调用相应的虚拟方法(Base:: VMethod之一),后者反过来调用模板方法实现(Impl::TMethod)。

只需要在派生实现(AImpl, BImpl)中实现模板方法TMethod,并使用derived <*Impl>。

class Base
{
public:
    virtual ~Base()
    {
    }

    template <typename T>
    T Method(T t)
    {
        return VMethod(t);
    }

private:
    virtual int VMethod(int t) = 0;
    virtual double VMethod(double t) = 0;
};

template <class Impl>
class Derived : public Impl
{
public:
    template <class... TArgs>
    Derived(TArgs&&... args)
        : Impl(std::forward<TArgs>(args)...)
    {
    }

private:
    int VMethod(int t) final
    {
        return Impl::TMethod(t);
    }

    double VMethod(double t) final
    {
        return Impl::TMethod(t);
    }
};

class AImpl : public Base
{
protected:
    AImpl(int p)
        : i(p)
    {
    }

    template <typename T>
    T TMethod(T t)
    {
        return t - i;
    }

private:
    int i;
};

using A = Derived<AImpl>;

class BImpl : public Base
{
protected:
    BImpl(int p)
        : i(p)
    {
    }

    template <typename T>
    T TMethod(T t)
    {
        return t + i;
    }

private:
    int i;
};

using B = Derived<BImpl>;

int main(int argc, const char* argv[])
{
    A a(1);
    B b(1);
    Base* base = nullptr;

    base = &a;
    std::cout << base->Method(1) << std::endl;
    std::cout << base->Method(2.0) << std::endl;

    base = &b;
    std::cout << base->Method(1) << std::endl;
    std::cout << base->Method(2.0) << std::endl;
}

输出:

0
1
2
3

注: Base::Method对于实际代码来说实际上是多余的(VMethod可以被设为public并直接使用)。 我添加它,使它看起来像一个实际的“虚拟”模板方法。

其他回答

我目前的解决方案如下(禁用RTTI -你也可以使用std::type_index):

#include <type_traits>
#include <iostream>
#include <tuple>

class Type
{
};

template<typename T>
class TypeImpl : public Type
{

};

template<typename T>
inline Type* typeOf() {
    static Type* typePtr = new TypeImpl<T>();
    return typePtr;
}

/* ------------- */

template<
    typename Calling
    , typename Result = void
    , typename From
    , typename Action
>
inline Result DoComplexDispatch(From* from, Action&& action);

template<typename Cls>
class ChildClasses
{
public:
    using type = std::tuple<>;
};

template<typename... Childs>
class ChildClassesHelper
{
public:
    using type = std::tuple<Childs...>;
};

//--------------------------

class A;
class B;
class C;
class D;

template<>
class ChildClasses<A> : public ChildClassesHelper<B, C, D> {};

template<>
class ChildClasses<B> : public ChildClassesHelper<C, D> {};

template<>
class ChildClasses<C> : public ChildClassesHelper<D> {};

//-------------------------------------------

class A
{
public:
    virtual Type* GetType()
    {
        return typeOf<A>();
    }

    template<
        typename T,
        bool checkType = true
    >
        /*virtual*/void DoVirtualGeneric()
    {
        if constexpr (checkType)
        {
            return DoComplexDispatch<A>(this, [&](auto* other) -> decltype(auto)
                {
                    return other->template DoVirtualGeneric<T, false>();
                });
        }
        std::cout << "A";
    }
};

class B : public A
{
public:
    virtual Type* GetType()
    {
        return typeOf<B>();
    }
    template<
        typename T,
        bool checkType = true
    >
    /*virtual*/void DoVirtualGeneric() /*override*/
    {
        if constexpr (checkType)
        {
            return DoComplexDispatch<B>(this, [&](auto* other) -> decltype(auto)
                {
                    other->template DoVirtualGeneric<T, false>();
                });
        }
        std::cout << "B";
    }
};

class C : public B
{
public:
    virtual Type* GetType() {
        return typeOf<C>();
    }

    template<
        typename T,
        bool checkType = true
    >
    /*virtual*/void DoVirtualGeneric() /*override*/
    {
        if constexpr (checkType)
        {
            return DoComplexDispatch<C>(this, [&](auto* other) -> decltype(auto)
                {
                    other->template DoVirtualGeneric<T, false>();
                });
        }
        std::cout << "C";
    }
};

class D : public C
{
public:
    virtual Type* GetType() {
        return typeOf<D>();
    }
};

int main()
{
    A* a = new A();
    a->DoVirtualGeneric<int>();
}

// --------------------------

template<typename Tuple>
class RestTuple {};

template<
    template<typename...> typename Tuple,
    typename First,
    typename... Rest
>
class RestTuple<Tuple<First, Rest...>> {
public:
    using type = Tuple<Rest...>;
};

// -------------
template<
    typename CandidatesTuple
    , typename Result
    , typename From
    , typename Action
>
inline constexpr Result DoComplexDispatchInternal(From* from, Action&& action, Type* fromType)
{
    using FirstCandidate = std::tuple_element_t<0, CandidatesTuple>;

    if constexpr (std::tuple_size_v<CandidatesTuple> == 1)
    {
        return action(static_cast<FirstCandidate*>(from));
    }
    else {
        if (fromType == typeOf<FirstCandidate>())
        {
            return action(static_cast<FirstCandidate*>(from));
        }
        else {
            return DoComplexDispatchInternal<typename RestTuple<CandidatesTuple>::type, Result>(
                from, action, fromType
            );
        }
    }
}

template<
    typename Calling
    , typename Result
    , typename From
    , typename Action
>
inline Result DoComplexDispatch(From* from, Action&& action)
{
    using ChildsOfCalling = typename ChildClasses<Calling>::type;
    if constexpr (std::tuple_size_v<ChildsOfCalling> == 0)
    {
        return action(static_cast<Calling*>(from));
    }
    else {
        auto fromType = from->GetType();
        using Candidates = decltype(std::tuple_cat(std::declval<std::tuple<Calling>>(), std::declval<ChildsOfCalling>()));
        return DoComplexDispatchInternal<Candidates, Result>(
            from, std::forward<Action>(action), fromType
        );
    }
}

我唯一不喜欢的是你必须定义/注册所有的子类。

我看了所有的14个答案,有些有原因,为什么虚拟模板的功能不能工作,其他人显示了一个工作周围。一个答案甚至表明虚类可以有虚函数。这不足为奇。

我的回答将给出一个直接的理由,为什么标准不允许虚模板函数。因为很多人都在抱怨。首先,我不敢相信有人说虚函数可以在编译时推导出来。这是我听过的最蠢的话。

不管怎样。我确定标准规定指向对象的this指针是其成员函数的第一个参数。

struct MyClass
{
 void myFunction();
}

// translate to
void myFunction(MyClass*);

既然我们都清楚了。然后,我们需要知道模板的转换规则。模板化的参数在它可以隐式转换的内容上受到极大的限制。我不记得所有的内容,但是你可以查看c++ Primer以获得完整的参考。例如,T*可转换为const T*。数组可以转换为指针。但是,派生类不能作为模板形参转换为基类。

struct A {};
struct B : A {};

template<class T>
void myFunction(T&);

template<>
void myFunction<A>(A&) {}

int main()
{
 A a;
 B b;

 myFunction(a); //compiles perfectly
 myFunction((A&)b); // compiles nicely
 myFunction(b); //compiler error, use of undefined template function
}

我希望你们能明白我的意思。你不能使用虚拟模板函数,因为就编译器而言,它们是两个完全不同的函数;作为隐式参数,此参数具有不同的类型。

虚拟模板不能工作的另一个原因同样有效。因为虚表是快速实现虚函数的最佳方式。

虚函数表

让我们从虚函数表及其工作原理的一些背景知识开始(来源):

[20.3] What's the difference between how virtual and non-virtual member functions are called? Non-virtual member functions are resolved statically. That is, the member function is selected statically (at compile-time) based on the type of the pointer (or reference) to the object. In contrast, virtual member functions are resolved dynamically (at run-time). That is, the member function is selected dynamically (at run-time) based on the type of the object, not the type of the pointer/reference to that object. This is called "dynamic binding." Most compilers use some variant of the following technique: if the object has one or more virtual functions, the compiler puts a hidden pointer in the object called a "virtual-pointer" or "v-pointer." This v-pointer points to a global table called the "virtual-table" or "v-table." The compiler creates a v-table for each class that has at least one virtual function. For example, if class Circle has virtual functions for draw() and move() and resize(), there would be exactly one v-table associated with class Circle, even if there were a gazillion Circle objects, and the v-pointer of each of those Circle objects would point to the Circle v-table. The v-table itself has pointers to each of the virtual functions in the class. For example, the Circle v-table would have three pointers: a pointer to Circle::draw(), a pointer to Circle::move(), and a pointer to Circle::resize(). During a dispatch of a virtual function, the run-time system follows the object's v-pointer to the class's v-table, then follows the appropriate slot in the v-table to the method code. The space-cost overhead of the above technique is nominal: an extra pointer per object (but only for objects that will need to do dynamic binding), plus an extra pointer per method (but only for virtual methods). The time-cost overhead is also fairly nominal: compared to a normal function call, a virtual function call requires two extra fetches (one to get the value of the v-pointer, a second to get the address of the method). None of this runtime activity happens with non-virtual functions, since the compiler resolves non-virtual functions exclusively at compile-time based on the type of the pointer.


我的问题,或者我是怎么来的

我尝试使用类似这样的东西,现在cubefile基类与模板优化加载函数,这将实现不同类型的立方体(一些存储像素,一些通过图像等)。

一些代码:

virtual void  LoadCube(UtpBipCube<float> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void  LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void  LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;

我想要它是什么,但它不会编译由于虚拟模板组合:

template<class T>
    virtual void  LoadCube(UtpBipCube<T> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
            long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;

我最终将模板声明移到了类级别。这种解决方案将迫使程序在读取数据之前了解它们将要读取的特定类型的数据,这是不可接受的。

解决方案

警告,这不是很漂亮,但它允许我删除重复的执行代码

1)在基类中

virtual void  LoadCube(UtpBipCube<float> &Cube,long LowerLeftRow=0,long LowerLeftColumn=0,
            long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void  LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
            long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;
virtual void  LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
            long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1) = 0;

2)和在儿童班

void  LoadCube(UtpBipCube<float> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }

void  LoadCube(UtpBipCube<short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }

void  LoadCube(UtpBipCube<unsigned short> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1)
{ LoadAnyCube(Cube,LowerLeftRow,LowerLeftColumn,UpperRightRow,UpperRightColumn,LowerBand,UpperBand); }

template<class T>
void  LoadAnyCube(UtpBipCube<T> &Cube, long LowerLeftRow=0,long LowerLeftColumn=0,
        long UpperRightRow=-1,long UpperRightColumn=-1,long LowerBand=0,long UpperBand=-1);

注意,LoadAnyCube没有在基类中声明。


下面是另一个堆栈溢出的答案: 需要一个虚拟模板成员解决方案。

至少在gcc 5.4中,虚函数可以是模板成员,但必须是模板本身。

#include <iostream>
#include <string>
class first {
protected:
    virtual std::string  a1() { return "a1"; }
    virtual std::string  mixt() { return a1(); }
};

class last {
protected:
    virtual std::string a2() { return "a2"; }
};

template<class T>  class mix: first , T {
    public:
    virtual std::string mixt() override;
};

template<class T> std::string mix<T>::mixt() {
   return a1()+" before "+T::a2();
}

class mix2: public mix<last>  {
    virtual std::string a1() override { return "mix"; }
};

int main() {
    std::cout << mix2().mixt();
    return 0;
}

输出

mix before a2
Process finished with exit code 0

在虚函数的情况下如何调用正确的函数?

虚表将包含类的每个虚函数的条目,在运行时,它将选择特定函数的地址,并调用各自的函数。

如何正确的函数必须被调用在虚拟情况下连同函数模板?

在函数模板的情况下,用户可以使用任何类型调用该函数。这里相同的函数根据类型有几个版本。现在,在这种情况下,对于同一个函数,由于版本不同,必须维护vtable中的许多项。