Question

Consider this code

class Foo {
private:
    Bar bar; //note: no reference

public:
   Foo(Bar& b) : bar(b) { }
};

Will Bar get copy-constructed?

Answer 1

That depends on the signatures of Bar‘s public constructors, either explicitly or implicitly defined.

To start with, the C++ standard allows for implicit conversion of references as long as the only difference in the underlying type is that the destination type is at least as cv-qualified than the source type, using the partial ordering defined in this table (C++11, §3.9.3/4):

no cv-qualifier < const
no cv-qualifier < volatile
no cv-qualifier < const volatile
const          < const volatile
volatile    < const volatile

So, taking that into account as well as §12.8/2:

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.

if Bar has a constructor with any of the following signatures:

Bar(Bar&);
Bar(Bar const&);
Bar(Bar volatile&);
Bar(Bar const volatile&);

then yes, b will be copy-constructed into Foo::bar.

---

EDIT: This was incorrect, I was thinking of operator= and the details of qualifying as a move-assignment operator.

Note that it’s possible to have an eligible constructor that is not a copy constructor:

Bar(Bar);

This will work (read: compile), but it is not technically a copy constructor.