When does the C++ default assignment operator become unaccessible?

Question

If I define an own assignment operator, which has a different signature than the normally generated default assignment operator:

struct B;
struct A {
void operator = (const B& b) {
    // assign something
}
};

does the default assignment operator, in this case operator = (A&) (or the like, correct me if wrong) become undefined/unaccessible?

AFAIK this is true for the default constructor, which doesn't exist, if we define some other constructor. But I am really not sure if this is the case for the other "magic" defaults.

The reason I ask: I want to avoid that the default copy constructor is accidently called via a implicit type conversion. If it doesn't exist, it could never happen.

Answer 1

No. 12.8/9 says that the assignment operator for class X must be non-static, non-template with a parameter of type X, X&, X const&, X volatile& or X const volatile&. And there is a note which emphasizes that the instantiation of a template doesn't suppress the implicit declaration.

Answer 2

Since A& operator=( B& ) has not the signature of A& operator=( const A& ), this does nothing to the synthesized assignement operator.

Take a look at this snippet at codepad.org - as far as an example counts as proof.

Test driving Comeau with it also shows that A& operator=( const A& ) is synthesized.