method chain a derived class method after calling

问题:

With a class and a derivative as shown below, is there a way for the base classes methods to return a object reference of the derived type instead of its own type so syntactically i can chain the methods together?

Suppose that object A has methods a1,a2 and derivative AD adds a method ad1 how would one go about making a valid method chain of AD_instance.a1().a2().ad1();?

Below are the two classes in question. Ignoring what it's actually doing, the method chaining is the only important part.

class AsyncWorker() {
pthread_t worker;
public:
  AsyncWorker();
  void *worker_run();
  AsyncWorker& Detach() { /*code*/ return *this;  }
  AsyncWorker& Start()  {
     pthread_create(&worker, NULL, &AsyncWorker::worker_helper, this);
     return *this;
  }
  static void *worker_helper(void *context) {
    return ((AsyncWorker *)context)->worker_run();
  }

};

class workerA : public AsyncWorker {
public: 
  int a;
  workerA(int i) { a = i; }
  void* worker_run() { ++a; sleep(1); }
  workerA& other_method_i_want_to_chain() { return *this };
};

Chained like so.

workerA A(0);
A.Start().Detach().other_method_i_want_to_chain();

回答1:

You could create a suitable overload in your derived class which dispatches to the base class version but return an object of itself:

class workerA {
    // ...
    workerA& Start() {
        this->AsyncWorker::Start();
        return *this;
    }
    workerA& Detach() {
        this->AsyncWorker::Detach();
        return *this;
    }
    // ...
 };

回答2:

Hope this makes the issue a little clearer for you.

#include <iostream>

struct Base
{
   virtual Base& foo() = 0;
};

struct Derived : Base
{
   virtual Derived& foo()
   {
      std::cout << "Came to Derived::foo()\n";
      return *this;
   }

   void bar()
   {
      std::cout << "Came to Derived::bar()\n";
   }

};

int main()
{
   Derived derived;
   derived.foo().bar(); // OK. From the interface of Derived, foo()
                        // returns a Derived&.
   Base& base = derived;
   base.foo().bar();    // Not OK. From the interface of Base, foo()
                        // returns a Base&.

   return 0;
}