include <iostream>

template<typename DTTYPE>
class Base
{
    public:
    Base(){};

        Base(const Base &bt, const DTTYPE &scale = 1.0){std::cout<<"base copy"<<std::endl;}

        Base(Base &&bt, const DTTYPE &scale = 1.0){std::cout<<"base move"<<std::endl;}
};

template<typename DTTYPE>
class Sed : public Base<DTTYPE>
{
    public:
    Sed(){};    
    Sed(const Sed &sd, const DTTYPE &scale = 1.0) : Base<DTTYPE>(sd, scale) {std::cout<<"copy sed"<<std::endl;}  
        Sed(Sed&& sd, const DTTYPE &scale = 1.0) : Base<DTTYPE>(sd, scale) {std::cout<<"move sed"<<std::endl;}    
};

int main(void)
{
    Sed<double> a;
    Sed<double> b(std::move(a));

    return 1;  
}

运行结果输出：   base copy
                 move sed

调用了基类的拷贝构造函数而不是移动构造函数。怎么才能调用基类的移动构造呢？

谢谢！
--
FROM 210.26.112.*

thx
【 在 MyWorkLife 的大作中提到: 】
: 这里的sd具名了以后变成了左值引用
:
--
FROM 219.246.59.*