C++ Template constructor, why is copy constructor being called?

Question

I have a class with a template constructor, and the code is actually calling the copy constructor, after the default-constructor, which does not make sense to me as the type is incorrect.

For example:

class A
{
  public:
    A(void); // default constructor
    A(const A& other); // copy constructor
    template<class TYPE>
    A(const TYPE& object_to_ref);  // template constructor
};

This template constructor works (is properly called in other cases), but is not properly recognized as the "correct" constructor from another template function:

template<class TYPE>
A& CreateA(const TYPE& object_to_ref)
{
  // THIS FUNCTION IS NEVER SPECIALIZED WITH "A", ONLY WITH "B" !!
  return *new A(object_to_ref);  // CALLS "A::A(const A&)" !!??
}

Example fail:

B my_b;
A& my_a = CreateA(my_b);  // "A::A(const A&)" called, not "A::A(const B&)"!

This does not make sense to me. The types are wrong to match the copy constructor. What happened? (MSVC2008)

My work-around is to not use the template constructor in this case:

template<class TYPE>
A& CreateA(const TYPE& object_to_ref)
{
  A* new_a = new A(); //DEFAULT CONSTRUCTOR
  new_a->setObjectToRef(object_to_ref); //OTHER TEMPLATE MEMBER FUNCTION
  return *new_a;
}

QUESTION: Why was the template constructor not called in this case?

(Work-around seems to behave properly, do you suggest an alternative?)

EDIT: B is unrelated, with no conversions specified between B and/or A:

class B
{
};

Tangential note: Code that does `return *new` is almost always a bad idea... — Oliver Charlesworth, Apr 02 '12 at 21:05
If `B` is implicitly convertible to `A` then this happens because the compiler finds a matching constructor without needing to ever generate the template code. — Chad, Apr 02 '12 at 21:08
http://stackoverflow.com/questions/752658/is-the-practice-of-returning-a-c-reference-variable-evil — Mat, Apr 02 '12 at 21:19
I can't reproduce that with GCC. Regardless of whether `B` inherits from `A`, the template constructor is called. — Philipp, Apr 02 '12 at 21:53

score 1 · Accepted Answer · answered Apr 02 '12 at 21:11

1

You didn't provide definition of B, so I'm going ahead assuming that A is B's ancestor and B can be implicitly cast to A. In this case your template for B is not being instantiated, because there is a perfectly suitable call already.

answered Apr 02 '12 at 21:11

cababunga

3,090
15
23

1

That doesn't make sense. You probably oversimplified your example. Is there any other implicit method of converting B to A? Like `operator A ()`? – cababunga Apr 02 '12 at 22:14
You were right: Didn't make sense, I oversimplified. There were more `template<>==>traits==>template<>` conversions going on, and one of them was unexpected-but-legitimate, ultimately causing the copy-constructor to be invoked in this case. My fault! (GOOD! IT WAS DRIVING ME CRAZY.) – charley Apr 03 '12 at 15:39

C++ Template constructor, why is copy constructor being called?

1 Answers1

Linked