Question

为什么在模板的情况下委托构造函数不起作用？复制构造函数不会在T=U的情况下调用常量复制构造函数，尽管没有template <typename U>和<U>，它可以正常工作。

template <typename T>
struct Class {
    Class () {
        std::cout << "Default constructor" << std::endl;
    }
    template <typename U>
    Class (const Class<U>& rhs) {
        std::cout << "Const copy constructor" << std::endl;
    }
    template <typename U>
    Class (Class<U>& rhs)
        : Class (const_cast<const Class<U>&> (rhs))
    {
        std::cout << "Copy constructor (templated)" << std::endl;
    }
/* // DOES NOT WORK WITHOUT THE NEXT:
    Class (Class& rhs)
        : Class (const_cast<const Class&> (rhs))
    {
        std::cout << "Copy constructor (not templated)" << std::endl;
    }
*/
};

Answer 1

请注意：模板构造函数永远不是（！）复制构造函数。将生成默认的复制构造函数（如果可能）。

struct NoCopy
{
    NoCopy() {}
    NoCopy(const NoCopy&) = delete;
};

template <typename T>
struct Test
{
    NoCopy member;
    Test() {};

    template <typename U>
    Test(const Test<U>&)
    {}
};

int main()
{
    Test<int> a;
    // The following error shows the missing generation of a default constructor,
    // due to the non copyable member. Without that member, the code compiles.
    // error: use of deleted function ‘Test<int>::Test(const Test<int>&)’
    Test<int> b(a);
}

委派构造函数和模板

1 个答案: