If pass-by-reference object is substiuted for variable in function, does it call copy constructor?

Question

I'm wondering if copy constructor is called or not. If the answer is "No", I don't know what happens. Could you tell me if you know answer for my question?

void func1(someClassA& obj_a)
{
 SomeClassB obj_b;
 obj_b.someClassA = obj_a; // this is the part I want to ask you!
 obj_b.parameter = something;
 func2(obj_b);
}

• SergeyA: What are all those classes?
• NathanOliver: This is assignment, not construction. Sounds like you could use a good C++ book
• Blood-HaZaRd: copy constructor is in invoked when you declare an object and you intialize it with already an exiting object like Obj B = Awith A an existing Obj var.
• Pierre Baret: Can't you define a copy constructor for someClass and try it by yourself?
• François Andrieux: @PierreBaret Proof by trial is dubious in c++. If you're testing something you aren't sure about, you have to keep the possibility of UB in the back of your mind.

Answer 1

In the line you commented, there is no difference when using someClassA& (reference) or someClassA (copy)

A reference is an alias [1] (another name) for the same variable. So you don't pass a pointer, and you don't copy the variable, but you send the reference-to-the-variable. You can use the reference just like the variable itself, because it IS actually referencing the original variable.

The commented line isn't affected by this at all. However the copy constructor is called when you call:

obj_x = someClassA(obj_a); // here, obj_a is given to the (copy-)constructor of someClassA.

What you have is an assignment like @eerorika said.

You can specify a custom assignment operator operator= and handle the action yourself.

---

More info about references/aliases and how they differ from pointers:

[1] Reference vs. pointer

• graziegrazie: > KYL3R Thank you so much for your detail answer.
• KYL3R: leave a like or accept the answer, please. If you need more details, don't hesitate to comment or extend your question post.
• graziegrazie: Thank you for information. I accepted the answer. I'm not used to use this site, so thank you.

Answer 2

No, that is an assignment expression. Copy assignment operator will be called. Assuming obj_b.someClassA is of type someClassA.

• eerorika: Assignment will be invoked, but not necessarily copy assignment.
• 463035818_is_not_an_ai: ah ok, sorry :)
• graziegrazie: > eerorika Thank you for your answer. You resolved my question!