Why doesn't the standard consider a template constructor as a copy constructor?

Question

Here's the definition of copy constructor, [class.copy.ctor/1]:

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments ([dcl.fct.default]).

Why does the standard exclude templates as copy constructors?

In this simple example, both constructors are copy constructors:

struct Foo {     Foo(const Foo &); // copy constructor     Foo(Foo &);       // copy constructor }; 

See this similar example:

struct Foo {      Foo() = default;       template <typename T>      Foo(T &) {          printf("here\n");      } };  int main() {     Foo a;     Foo b = a; } 

In this example, here will be printed. So it seems that my template constructor is a copy constructor, at least it behaves like one (it is called in a context where copy constructors are usually called).

Why is the "non-template" requirement there in the text?

Answer 1

Let's put templates aside for a second. If a class doesn't declare a copy constructor, an implicitly defaulted one is generated. It may be defined as deleted, but it's defaulted nonetheless.

A member template is not a member function. Members are instantiated from it only when needed.

So how can a compiler know from the class definition alone whether or not a specialization with T = Foo will ever be needed? It can't. But it's exactly that on which it needs to base a decision of how to handle a potential need for an implicitly defaulted copy constructor (AND move constructor). That becomes messy.

The easiest approach is to exclude templates. We'll always have some copy constructor anyway, it will do the correct thing^TM by default, and will be favored by overload resolution because it's not instantiated from a template.