在deques之间移动唯一指针

Question

我有一些指向对象的唯一指针的容器类。

我现在需要接受其中的一个，在不同的课堂上对他们做些什么，然后交给另一个班级（并在某个时间点回收它们并将其放回到容器中）。

下面是代码的概要。但是我收到用正确的方式来迷惑独特的移动与指针功能：

using uptrPod = unique_ptr<Pod>; 
using dqUptrPods = deque<uptrPods>; 

class Container { 
public: 
    Container() : 
     n_elements_(500) 
    { 
     for (auto i = 0; i < n_elements_; i++) { 
     free_pods_.push_back(uptrPod(new Pod())); 
    } 

    const uptrPod&& getPod() { 
     auto up_pod= std::move(free_pods_[0]); 
     free_pods_.erase(free_pods_.begin()); 
     return up_pod; 
    } 

    void returnPod(const uptrPod&& up_pod) { 
     free_pods_.push_back(up_pod); 
    } 

private: 
    long n_elements_; 
    dqUptrPods free_pods_; 
}; 

class PodTracker { 

    void addPod(const uptrPod&& up_pod) { dq_pods_.pushback(up_pod); } 
    dqUptrPods dq_pods_; 
}; 

class PodHandler { 

    void movePod() { 
     up_pod = std::move(container_->getPod()); 

     /// do stuff with pod 

     pod_tracker_->addPod(up_pod); 
    } 

    Container* container_; 
    PodTracker* pod_tracker_; 
}; 

我收到错误：

cannot bind std::unique_ptr l value to const uptrPod&& { aka const std::unique_ptr &&

我如何用手绕类之间的指针？

Answer 1

通/按值返回std::unique_ptr实例，并明确使用std::move：

uptrPod getPod() { 
    auto up_pod= std::move(free_pods_[0]); 
    free_pods_.erase(free_pods_.begin()); 
    return up_pod; 
} 

void returnPod(uptrPod up_pod) { 
    free_pods_.push_back(std::move(up_pod));   
} 

class PodTracker{ 
    void addPod(uptrPod up_pod) { dq_pods_.push_back(std::move(up_pod)); } 
    dqSptrPods dq_pods_; 
}; 

void movePod() { 

    // `std::move` is not necessary here, as `getPod()` 
    // is already a temporary (rvalue) 
    auto up_pod = container_->getPod(); 

    /// do stuff with pod 

    pod_tracker_->addPod(std::move(up_pod)) ; 
} 

example on wandbox

---

注意const右值引用没有多大意义 - 右值引用。

The main purpose of rvalue references is to allow us to move objects instead of copying them. And moving the state of an object implies modification. As a result, the canonical signatures for the move constructor and the move assignment operator both take its argument as a non-const rvalue reference.

（来源："What are const rvalue references good for?"）