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

Answer 1

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

Given it were different and copy constructors could be templates. How could a non-copy constructor not be ambiguous in the presence of a copy constructor template? Consider this:

struct Foo {
   // ctor template: clearly useful and necessary
   template 
      Foo(const T&) {}

   // copy ctor: same signature! can't work
   template 
      Foo(const T &) {}
};

Besides, constructing a Foo from an object that is not a Foo can be achieved by either conversion or ordinary construction, but allowing for copy-construction from a non-Foo object changes the notion of copying to copying including conversion. But this can already be implemented with the existing scheme (conversion or non-copy construction).

In this example, here will be printed. So it seems that my template constructor is a copy constructor

The example that you show doesn't invoke copy construction, but an ordinary, implicit construction. If you change the constructor template to

template 
Foo(const T &) {
//  ^^^^^
    printf("here\n");
}

then Foo b = a; results in the compiler-generated copy constructor being called. Note that the copy ctor generated by the compiler has this signature:

Foo(const Foo&);

This requires adding a const-qualifier to a in Foo b = a;. The original constructor template Foo(T&) in your snippet is a better match, as no const-qualifier is added.