复制模板类的构造函数

Question

I read that template copy-con is never default copy onstructor, and template assignment-op is never a copy assignment operator.

I couldn't understand why this restriction is needed and straight away went online to ideone and return a test program but here copy constructor never gets called on further googling I came across templatized constructor and tried that but still it never calls copy constructor.

#include <iostream>
using namespace std;

template <typename T> class tt
{
    public :
    tt()
    {
        std::cout << std::endl << "   CONSTRUCTOR" << std::endl;
    }
    template <typename U> const tt<T>& operator=(const tt<U>& that){std::cout << std::endl << "   OPERATOR" << std::endl;}
    template <typename U> tt(const tt<U>& that)
    {
        std::cout << std::endl << "    COPY CONSTRUCTOR" << std::endl;
    }
};


tt<int> test(void)
{
    std::cout << std::endl << "      INSIDE " << std::endl; tt<int> a; return a;
}

int main() {
    // your code goes here
    tt<int> a ; a = test();

    return 0;
}

Can someone explain me the whole reason behind putting this restriction and also how to write a copy constructor of template class.

Thanks

Answer 1

There are strict rules what constitutes a copy constructor (cf. C++11, 12.8):

• It is not a template.

• For a class T , its first argument must have type T & or T const & or T volatile & or T const volatile & .

• If it has more than one argument, the further arguments must have default values.

If you do not declare a copy constructor, a copy constructor of the form T::T(T const &) is implicitly declared for you. (It may or may not actually be defined, and if it is defined it may be defined as deleted.)

(The usual overload resolution rules imply that you can have at most four copy constructors, one for each CV-qualification.)

There are analogous rules for move constructors, with && in place of & .

Answer 2

I can't comment on why this is how it is, but here's how you write a copy constructor and assignment operator for a class template:

    template <class T>
    class A
    {
      public:
        A(const A &){}
        A & operator=(const A& a){return *this;}
    };

and that's it.
The trick here is that even though A is a template, when you refer to it inside the class as A (such as in the function signatures) it is treated as the full type A<T> .