c++ template copy constructor on template class

问题:

I have a template class that has a template copy constructor. The problem is when I instantiate this class using another instance of this class with the same template type, my template copy constructor is not called. Why doesn't it match ? Here is the code snippet:

#include <iostream>

template <typename T>
class MyTemplateClass
{
    public:
        MyTemplateClass()
        {
            std::cout << "default constructor" << std::endl;
        }

        /*
        MyTemplateClass(const MyTemplateClass<T>& other)
        {
            std::cout << "copy constructor" << std::endl;
        }
        */

        template <typename U>
        MyTemplateClass(const MyTemplateClass<U>& other)
        {
            std::cout << "template copy constructor" << std::endl;
        }
};

int main()
{
    MyTemplateClass<int> instance;
    MyTemplateClass<int> instance2(instance);
    return EXIT_SUCCESS;
}

The output is

default constructor

But if I explicitly write the default copy constructor (by uncommenting it), then the output becomes

default constructor
copy constructor

I really don't get it. I tested it with my local compiler (clang-500.2.79) and with this one (gcc 4.9.2) and got the same result.

回答1:

A copy constructor is of the form X(X& ) or (X const&), and will be provided for you by the compiler if you didn't declare one yourself (or a few other conditions which are not relevant here). You didn't, so implicitly we have the following set of candidates:

MyTemplateClass(const MyTemplateClass&);
template <typename U> MyTemplateClass(const MyTemplateClass<U>&);

Both are viable for

MyTemplateClass<int> instance2(instance);

Both take the same exact arguments. The issue isn't that your copy constructor template doesn't match. The issue is that the implicit copy constructor is not a function template, and non-templates are preferred to template specializations when it comes to overload resolution. From [over.match.best], omitting the unrelated bullet points:

Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICS_i(F1) is not a worse conversion sequence than ICS_i(F2), and then
— [...]
— F1 is not a function template specialization and F2 is a function template specialization, or, if not that,
— [...]

That's why it calls your implicit (and then, your explicit) copy constructor over your constructor template.

回答2:

When you do not have a copy constructor in you code, the compiler will implicitly generate it. Therefore when this line is executed:

MyTemplateClass<int> instance2(instance);

A copy constructor is being executed, though obviously not yours. I think that templating has nothing to do with it.

Read more about it here: Implicitly-defined copy constructor

回答3:

I think REACHUS is right, the compiler is generating a default copy-constructor (as it would with a non-template class too) and preferring this over your template as it's more specialised. You should make your "normal" copy-constructor private, or better, use the C++11 'deleted' keyword to mark the function as unusable. EDIT: This doesn't compile, sorry, I wasn't able to test it at the time.