Question

My current implementation uses lots of copy constructors with this syntax

MyClass::Myclass(Myclass* my_class)

Is it really (functionnaly) different from

MyClass::MyClass(const MyClass& my_class)

and why?

I was adviced that first solution was not a true copy constructor. However, making the change implies quite a lot of refactoring.

Thanks!!!

Answer 1

It’s different in the sense that the first isn’t a copy constructor, but a conversion constructor. It converts from a MyClass* to a MyClass.

By definition, a copy-constructor has one of the following signatures:

MyClass(MyClass& my_class)
MyClass(const MyClass& my_class)
//....
//combination of cv-qualifiers and other arguments that have default values

12.8. Copying class objects

2) 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 (8.3.6).113) [ Example: X::X(const
X&) and X::X(X&,int=1) are copy constructors.

EDIT: you seem to be confusing the two.

Say you have:

struct A
{
   A();
   A(A* other);
   A(const A& other);
};

A a;     //uses default constructor
A b(a);  //uses copy constructor
A c(&a); //uses conversion constructor

They serve different purposes alltogether.