在没有开销的情况下实现 push_back 的最佳方法是什么

Question

I'm trying to implement a Queue where you pass to it an Object to be added to the Queue.

struct Node {
    T data;
    Node *next, *prev;
};    
// Push data to the back of the list.
template <class T> T& CircularQueue<T>::push_back(const T&& new_data)
{
    Node* new_node = new Node();
    new_node->data = std::move(new_data);
    link_node(new_node, m_head);
    return new_node->data;
}

The problem with my current approach is there is too much overhead (as i come from C these things bothers me). for example image i will add an object from MyClass:

CircularQueue<MyClass> list;
list.push_back(MyClass(arg1, arg2));

The first problem is that the MyClass needs to have a constructor without argument to be used in Node* new_node = new Node(); since creating a Node structure will call the constructor of the object inside it which is MyClass. i tried with std::vector and it didn't require this.

The second problem is too much overhead, list.push_back(MyClass(arg1, arg2)); will create an rvalue object in the stack then send to push_back , it then creates a new object (with no argument list) in the heap then move all of its members to the new object using move assignment, is there any faster solution?

Answer 1

you can emplace_back your Node

template <class T> 
class CircularQueue {
    template<typename... U>
    T &emplace_back(U&&... u)
    {
       Node *new_node = new Node{{std::forward<U>(u)...}}; // <data is created here
        // link_node(new_node, m_head);
       return new_node->data;
    }
};
void foo() {
    CircularQueue<Data> x;
    // Do not create a Data, pass the parameters you need to create
    x.emplace_back(10, 20);
    // If you actually need to, you can of course copy or move an existing Data
    Data y(20, 30);
    x.emplace_back(y); // copies y
    x.emplace_back(std::move(y)); // moves y
}

https://godbolt.org/z/z68q77