使用unique_ptr作为成员的C++模板的构造函数失败

Question

当类模板包含另一个类的unique_ptr时,该类的构造函数不会将unique_ptr移动到新的Object中.使用相同的类,但没有模板,构造函数按预期生成对象.

#include <iostream>
class test1{
public:
    std::string s_;
    test1(std::string s):s_(s){};
};
class testu{
public:
    std::unique_ptr<test1> us_;
    testu(std::unique_ptr<test1> us):us_(std::move(us)){};
};

template <int i>
class testt {
public:
    std::unique_ptr<test1> us_;
    testt<i>(std::unique_ptr<test1> us):us_(std::move(us)){};
};

template class testt<0>;

int main(int argc, const char * argv[]) {
    //without template
    std::unique_ptr<test1> us(new test1("test"));
    testu* t1=new testu(move(us));
    std::cout<<t1->us_->s_<<"\n";

    //with template the constructor fails!
    std::unique_ptr<test1> ust(new test1("test"));
    testt<0>* t2=new testt<0>(std::move(us));
    std::cout<<t2->us_->s_<<"\n";  //crash!
    return 0;
}

Answer 1

这只是一个错字:

testt<0>* t2 = new testt<0>(std::move(us));

这条线应该是

testt<0>* t2 = new testt<0>(std::move(ust));

us已经从第一部分移开main,因此倒数第二行的访问无效:

std::cout<<t2->us_->s_<<"\n"; 
//             ^^^
//              | Was already moved from, access causes UB

修复使程序运行正常.