c++ template copy constructor not available or explicit - c++

Question

I wrote a code with template copy constructor to get a better understanding of the concept as I am new to it but the following code fails to compile
#include <iostream>
#include <vector>
using namespace std;
template <typename T>
class Grid
{
public:
explicit Grid(size_t inWidth = kDefaultWidth, size_t inHeight = kDefaultHeight);
virtual ~Grid();
template <typename E>
Grid(const Grid<T>& src);
static const size_t kDefaultWidth = 10;
static const size_t kDefaultHeight = 10;
std::vector<std::vector<T>> mCells;
size_t mWidth, mHeight;
};
template <typename T>
template <typename E>
Grid<T>::Grid(const Grid<T>& src)
{
cout << "Copy constructor working " << endl;
}
int main()
{
Grid<double> myDoubleGrid;
Grid<double> newDoubleGrid(myDoubleGrid);
return 0;
}
On compiling the above code in visual studio following error appears:-
ERROR:-
Severity Code Description Project File Line Suppression State
Error C2558 class 'Grid': no copy constructor available or copy constructor is declared 'explicit'
I have tried to replace the argument's template with E and It shows more errors (strange ones)
template <typename T>
template <typename E>
Grid<T>::Grid(const Grid<E>& src)
{
cout << "Copy constructor working " << endl;
}
Errors:
Severity Code Description Project File Line Suppression State
Error LNK2019 unresolved external symbol "public: __thiscall Grid::Grid(unsigned int,unsigned int)" (??0?$Grid#N##QAE#II#Z) referenced in function _main
Error LNK1120 2 unresolved externals
Error LNK2019 unresolved external symbol "public: virtual __thiscall Grid::~Grid(void)" (??1?$Grid#N##UAE#XZ) referenced in function "public: virtual void * __thiscall Grid::`scalar deleting destructor'(unsigned int)" (??_G?$Grid#N##UAEPAXI#Z)

Answer 1

A template constructor is never(!) a copy constructor. The constructor in your code is a converting constructor, only.
You may want both:
#include <iostream>
#include <vector>
// Please do not have this ugliness in a header!
using namespace std;
template <typename T>
class Grid
{
public:
explicit Grid(size_t inWidth = kDefaultWidth, size_t inHeight = kDefaultHeight)
// initialize members ...
{}
// Without this copy-constructor the compiler generates a copy-constructor
// (with no output, of course)
Grid(const Grid& src)
// initialize members ...
{
cout << "Copy constructor" << endl;
}
template <typename E>
Grid(const Grid<E>& src)
// initialize members ...
{
cout << "Converting constructor" << endl;
}
static const size_t kDefaultWidth = 10;
static const size_t kDefaultHeight = 10;
size_t mWidth, mHeight;
};
int main()
{
Grid<double> myDoubleGrid;
Grid<double> newDoubleGrid(myDoubleGrid); // Copy constructor
Grid<int> newIntGrid(myDoubleGrid); // Converting constructor
return 0;
}
See also:
http://en.cppreference.com/w/cpp/language/copy_constructor
http://en.cppreference.com/w/cpp/language/converting_constructor
Derive from template constuctor of template base class

Answer 2

Note that Grid<T>(Grid<U> const&) is not technically a copy constructor, it is a converting constructor. The compiler still generates a copy constructor for you.

Answer 3

The thing to which you call as dangerous error is happening because of your silly mistakes:-
1) You have provided declarations for both constructors and destructors. And as you are saying that it is your one and only source file. It means that you must have not defined them and which is the reason for the linker errors that you are getting. Sometimes these pretty simple problems can prove to be very dangerous and time consuming.
You could have replaced the semicolon with a pair of braces if you don't want to provide any detailed implementation.
2) The thing that you are trying to do is completely logicless as you are creating a converter constuctor with the ssignature similar to copy constructor. Anyway compiler will give preference to copy constructor than your converter constructor.
Ultimately your definition for the constructor will never actually be called.