在下面的 :
template<typename Derived>
class Base:
{
inline Derived& operator=(const Base<Derived>& x);
}
此声明是否删除了默认的复制赋值运算符或者我有两个运算符:
inline Derived& operator=(const Base<Derived>& x);
// (declared by me)
和
inline Base<Derived>& operator=(const Base<Derived>& x);
// (declared by the compiler)
在这种情况下,当我调用函数时,编译器将如何获得正确的运算符?