Question

In C++, if I define a copy constructor and operator= that take a non-const reference to the class, is the compiler supposed to still supply default versions for const reference?

struct Test {
  Test(Test &rhs);
  Test &operator=(Test &rhs);

private:
  // Do I still need to declare these to avoid automatic definitions?
  Test(const Test &rhs);
  Test &operator=(const Test &rhs);
};

Answer 1

No, if you define a copy constructor and assignment operator, the compiler will not implicitly declare or define it’s own. Note that the definition of copy-constructor allows for the argument to be taken by either const or non-const reference, so your constructor is indeed a copy-constructor. Similarly for operator=

[Omitting a big part of the details, in particular under what circumstances the implicitly declared special member functions will also be implicitly defined]

12.8 [class.copy]/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).

12.8 [class.copy]/7 If the class definition does not explicitly declare a copy constructor, one is declared implicitly.

12.8 [class.copy]/17 A user-declared copy assignment operator X::operator= is a non-static non-template member function of class X with exactly one parameter of type X, X&, const X&, volatile X& or const volatile X&.

12.8 [class.copy]/18 If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly.