C++: error when have private copy ctor with public assignment operator

Question

Hi guys,

Can one of you explain why the following piece of code does not compile?

#include <iostream>

using namespace std;

class Foo
{
public:
  Foo() { cout << "Foo::Foo()" << endl << endl; }
  Foo& operator=(const Foo&) { cout << "Foo::operator=(const Foo&)" << endl << endl; }
private:
  Foo(const Foo& b) { *this = b; cout << "Foo::Foo(const Foo&)" << endl << endl; }
};

int main()
{
  Foo foo;

  foo = Foo();
}

The error I receive:

$ g++ -o copy_ctor_assign copy_ctor_assign.cc && ./copy_ctor_assign
copy_ctor_assign.cc: In function 'int main()':
copy_ctor_assign.cc:10: error: 'Foo::Foo(const Foo&)' is private
copy_ctor_assign.cc:17: error: within this context

Note: when I remove the private: keyword the code compiles but the copy ctor is never called. So why does it err when it's private?

Not sure if it's important but I'm using:

$ g++ --version
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-44)
Copyright (C) 2006 Free Software Foundation, Inc.

Answer 1

Copy Ctor is called when:

1. passing an object by value as parameter to a function,
2. returning an object from a function.

So you are certainly doing one or both of these case somewhere in your code. You should set Copy Ctor as public or avoid the 2 previous cases.

Answer 2

Copy constructor would be called if you write

Foo foo; // normal constructor
Foo foo1(foo); //copy constructor

In your case, first the default constructor is called and then the operator= method.

Answer 3

Assuming that the code you've posted is the only code in the project, and there's no covert passing of Foos by value going on anywhere, all I can figure is that gcc is optimizing

Foo foo;
foo = Foo();

to

Foo foo = Foo();

...which is unsound, as the first form is a default-construct and an assignment, while the second is equivalent to

Foo foo(Foo());

...which is clearly a copy construction. If I'm right, the copy constructor is not being run because GCC can optimize away the redundant temporary; this is permitted by the C++ spec.

In general, it is not a good idea to have assignment operators and copy constructors at different protection levels; as you've seen, the results can be unintuitive.

Answer 4

That code compiles with gcc 4.3.3 and 4.4.1. Maybe that's just a bug in gcc 4.1?

Answer 5

#include <iostream>

using namespace std;

class Foo
{
public:
  Foo() { cout << "Foo::Foo()" << endl << endl; }
  Foo& operator=(const Foo&) { cout << "Foo::operator=(const Foo&)" << endl << endl; }
  Foo(const Foo& b) { *this = b; cout << "Foo::Foo(const Foo&)" << endl << endl; }
};

int main()
{
  Foo f1;// default constructor called

  Foo f2 = f1; //copy constructor called
}

Check this, in Foo f2=f1; ( f2 is created using copy constructor)

Answer 6

You are initializing a reference from temporary.
The standard states:
The temporary should be initialized (8.5.3 par 5)"using the rules for a non-reference copy initialization (8.5)".

The copy construction is removed for the temporary (permitted by the standard. 12.8 par 5).
However, the standard clearly states (12.2 par 1):
"Even when the creation of the temporary object is avoided (12.8), all the semantic restrictions must be respected as if the temporary object was created. [Example: even if the copy constructor is not called, all the semantic restrictions, such as accessibility (clause 11), shall be satisfied. ]"

(also, when looking for the right quote, found this duplicate :)

Edit: adding relevant location from the standard