C++ 模板是否继承另一个模板？

我试着从大到小排序 我的数学向量继承自向量 现在的问题是我看不到inhert函数（见下文&&&&&&）

模板
类向量
{
受保护的：
T*阵列；
整数大小；
国际能力；
公众：
向量（）；
向量（向量和拷贝数组）；
T运算符=（常数T*元素）；
void运算符=（常量T元素）；
T*运算符[]（常数int索引）常数；
ostream&operator是的，它是从
向量
继承的正确方法。例如，使用公共方法write

template<class T>
void MathVector<T>::Sort(const int Direction)
{
    Vector<T>::yourNonPrivateMethod();
    auto test = Vector<T>::yourNonPrivateMember;
};

编辑
下面是一个简化的示例：

#include <iostream>

template<typename T> class Base
{
public:
    Base() : member( static_cast<T>( 0 ) ) {};

    virtual ~Base() = default;

    T member;

    void baseMemberFunction()
    {
        std::cout << "Base<T>::memberFunction()" << std::endl;
    }

    void anotherBaseMemberFunction()
    {
        std::cout << "Base<T>::anotherBaseMemberFunction()" << std::endl;
    }
};


template<typename T> class Derived : public Base<T>
{
public:
    Derived() : Base<T>(), member(1) {}

    T member;

    using Base<T>::anotherBaseMemberFunction;

    void derivedMemberFunction()
    {
        std::cout << "Derived<T>::derivedMemberFunction()" << std::endl;
        Base<T>::baseMemberFunction();
        anotherBaseMemberFunction();
        std::cout << "Derived<T>::member = " << member << std::endl;
        std::cout << "Base<T>::member = " << Base<T>::member << std::endl;
    }
};


int main()
{
    Derived<int> derived;

    std::cout << " --- Call baseMemberFunction --- " << std::endl;
    derived.baseMemberFunction();

    std::cout << " --- Call derivedMemberFunction --- " << std::endl;
    derived.derivedMemberFunction();

    return 0;
}

#包括
模板类基
{
公众：
Base（）：成员（static_cast（0））{}；
virtual~Base（）=默认值；
T成员；
void baseMemberFunction（）
{
（我不知道为什么你的方式和我的方式可能与我错过的某件事有关wirte@yarin_hala：请检查编辑的代码。使用
g++-5-std=c++11-Wall-pedantic
编译。它应该准确地映射您的问题。顺便说一下：使基类中的析构函数成为虚拟的。
template<class T>
void MathVector<T>::Sort(const int Direction)
{

this->  &&&&&&/* i can't see the public mathods of vector*/&&&&&&


};

template<class T>
void MathVector<T>::Sort(const int Direction)
{
    Vector<T>::yourNonPrivateMethod();
    auto test = Vector<T>::yourNonPrivateMember;
};

using Vector<T>::yourNonPrivateMethod;

#include <iostream>

template<typename T> class Base
{
public:
    Base() : member( static_cast<T>( 0 ) ) {};

    virtual ~Base() = default;

    T member;

    void baseMemberFunction()
    {
        std::cout << "Base<T>::memberFunction()" << std::endl;
    }

    void anotherBaseMemberFunction()
    {
        std::cout << "Base<T>::anotherBaseMemberFunction()" << std::endl;
    }
};


template<typename T> class Derived : public Base<T>
{
public:
    Derived() : Base<T>(), member(1) {}

    T member;

    using Base<T>::anotherBaseMemberFunction;

    void derivedMemberFunction()
    {
        std::cout << "Derived<T>::derivedMemberFunction()" << std::endl;
        Base<T>::baseMemberFunction();
        anotherBaseMemberFunction();
        std::cout << "Derived<T>::member = " << member << std::endl;
        std::cout << "Base<T>::member = " << Base<T>::member << std::endl;
    }
};


int main()
{
    Derived<int> derived;

    std::cout << " --- Call baseMemberFunction --- " << std::endl;
    derived.baseMemberFunction();

    std::cout << " --- Call derivedMemberFunction --- " << std::endl;
    derived.derivedMemberFunction();

    return 0;
}