使用不带括号的operator()()？

Question

以下代码是对象包装器的简化版本。我希望能够Object无缝访问底层，即不需要括号，如评论所述：

struct A
{
  void Func() {}
};

template <typename Object>
struct ObjectWrapper
{
  ObjectWrapper(Object& o) : object_(&o) {}

  operator Object& () { return *object_; }
  Object& operator ()() { return *object_; }

  Object* object_;
};


int main()
{
  A a;
  ObjectWrapper<A> obj(a);

  //
  // Trying to call Func() on the A object that 'obj' wraps...
  //

  obj.operator A& ().Func(); // Messy

  obj().Func(); // Better but still has parentheses

  // I really want to be able to say this:
  // obj.Func()
  // ...but cannot see how!
}

谁能建议一种方法来做到这一点？

Answer 1

我认为你需要重载运算符->和/或*（这就是智能指针的完成方式）：

template <typename Object>
struct ObjectWrapper {
    ObjectWrapper(Object& o)
        : object_(&o)
    {
        LOG();
    }

    Object* operator->() const
    {
        LOG();
        return object_;
    }

    Object& operator*() const
    {
        LOG();
        return *object_;
    }

    Object* object_;
};

int main()
{
    A a;
    ObjectWrapper<A> obj { a };
    obj->Func();
    (*obj).Func();
}

https://godbolt.org/z/ErEbxWE4P