引用類型的初始化無效？

Question

怎麼回事？這是它引用的功能。我試圖讓這個作爲一個拷貝構造函數

template <class T> 
const queue<Base>& queue<T>::operator=(const queue<Base> &q){ 
// Doesn't need to copy if they are the same object 
if (this != &q){ 
    delete [] data; 

    length = q.length; 
    capacity = q.capacity; 
    front = q.front; 

    data = new T[capacity]; 

    for (int i = 0; i < capacity; i++){ 
     data[i] = q.data[i]; 
    } 
} 

return this; 
} 

Answer 1

工作，這是你的錯誤

return this; 

this是一個指針。您的operator =被聲明爲返回參考。指針不能轉換爲引用。這是錯誤信息告訴你的。