why does --list.end() compile?

Question

• list's end() returns a copy of the past-the-end iterator, right?
• Therefore, list.end() is an rvalue, right?
• the -- operator-function overloaded for list iterator takes a non-const reference, right?
• you can't bind rvalues to non-const references, right?

So how come does

std::list<int> lst;
// ... 
--l.end();` 

compile?

As correctly pointed out, my third point is not necessarily right. But then how about this code which also compiles?

struct A{};

void f(A&)
{

}
A a()
{
    return A();
}

int main()
{
    f(a());
}

Answer 1

• the -- operator-function overloaded for list iterator takes a non-const reference, right?

This point is wrong. The operator-- is a member function, and a member function can be invoked on a temporary. You don't pass any object to this member function as argument. so the question of binding rvalue to non-const reference doesn't arise in the first place.

---

As for the edit (passing rvalue to f(A&){}), it is a non-standard extension. I guess you're using Microsoft Compiler, because I know it has this non-standard extension which is stupid in my opinion.