Question

该表达式无法分配什么？我正在尝试在C ++中实现双向链表。这是节点的struct：

struct NodoLDL
{
   T dato;
   NodoLDL *anterior;
   NodoLDL *siguiente;

   NodoLDL(const T &elem, NodoLDL *ant = nullptr, NodoLDL *sig = nullptr):
      dato(elem),
      anterior(ant),
      siguiente(sig)
   {}
};

这是列表类：

template <typename T>
class LDL
{
private:
   #include "nodoldl.h"
   size_t listSize;
   NodoLDL *listFront; //head
   NodoLDL *listBack; //tail

public:
   LDL() : listSize(0), listFront(nullptr), listBack(nullptr)
   {}
   void pop_front() const;
(. . .)
}

这是pop_front函数，我在listFront = temp;上收到一条错误消息，该代码的含义是在每一行中作为注释：

template<typename T>
void LDL<T>::pop_front() const
{
    if(empty()){
        throw invalid_argument("pop_front() on empty list");
    }else{
        NodoLDL *temp = listFront->siguiente; 
        ///temp points to the following node in the list (from listFront)

        temp->anterior = nullptr; 
        //Now that temp is at the "second" node in the list,
        //the pointer to the node before will be null, 
        //as the actual listFront will be deleted.

        delete listFront;
        //It deletes the node

        listFront = temp;
        //Now listFront is equal to temp, which is at the second node
    }
}

为什么这种逻辑是错误的？我该如何解决？

Answer 1

该函数更改调用它的对象的状态。使其成为const成员函数没有任何意义。将其更改为非{const成员函数。

pop_front（），错误为“在只读对象中分配成员”

1 个答案: