链表分段故障11

Question

我的一个作业有问题。我需要做一个链表连接，但重新定义运算符+后，我有分割错误。下面是代码:

在main.cpp:

case CONCAT:
        do
    {
      std::cout << "Which two list do you want to concat?(1-3) ";
      std::cin >> s;
            std::cin >> g;
            selectedlist =atoi(s.c_str());
            selectedlist2 =atoi(g.c_str());
    } while ((
                 selectedlist== 0 && 
                 s != "0" && 
                 (selectedlist > 3 || selectedlist < 1)
             ) && 
             (
                  selectedlist2== 0 && 
                  g != "0" && 
                  (selectedlist2 > 3 || selectedlist2 < 1)
             )
            );
        Lista[selectedlist-1]+Lista[selectedlist2-1];
        std::cout<<"Ready";
   break;

头文件:

class lista
{
    public:
        enum Exceptions{EMPTY};
        lista() : first(NULL),last(NULL),current(NULL){ first = new Elem(0,0);}
        virtual ~lista();
        lista(const lista& s);
        int Current() const {return current->Adat;}
        void First()   {current = first->next;}
        bool End()     const {return current==NULL;}
        void Next()    {current = current->next;}
        void Vegere(int e);
        void Elejerol();
        bool Urese();
        int Eleje();
        friend std::ostream& operator<<(std::ostream& s, const lista& a);
        friend lista operator+(lista& a, lista& b);
    private:
        struct Elem{
                    int Adat;
                    Elem* next;
                    Elem(int c, Elem* n): Adat(c), next(n){};
                    };
        Elem* first;
        Elem* last;
        Elem* current;
};

在list .cpp:

lista operator+(lista& a, lista& b)
{
  if(b.first->next!=NULL && a.first->next!=NULL)
    {
        a.last->next = b.first->next;
        a.last = b.last;
        b.first = new lista::Elem(0,0);
        b.last = NULL;
        b.current = NULL;
    }
   else
   {
        throw lista::EMPTY;
   }
    return a;
}
void lista::Vegere(int e) { 
  Elem* p = new Elem(e,0); 
  if(last==NULL) { 
     first -> next = p; 
     last = p; 
  } 
  else { 
    last -> next = p; 
    last = p; 
  } 
} 

它抱怨，所有其他函数(空，加数字等)工作正常。我做错了什么?

void lista::Vegere(int e) 
{ 
    Elem* p = new Elem(e,0); 
    if(last==NULL) 
    { 
        first -> next = p; last = p; 
    } 
    else 
    { 
        last -> next = p; last = p; 
    } 
}

析构函数:

lista::~lista()
{
    Elem *p, *v;
    p = first;
    v = first -> next;
    while( v!=NULL)
    {
        delete p;
        p = v;
        v = v -> next;
    }
    delete p;
}

我已经弄明白了。我没有在这里发布这部分代码，但经过一些调试后，我得到了它。

错误如下:

    lista::lista(const lista& s){
        if(s.first->next==NULL)
        {
            first->next = last = NULL;
        }
        else
        {
        Elem* q = new Elem(s.first->Adat,NULL);
        first = q;
        for(Elem* p=s.first->next;p!=NULL;p=p->next)
        {
            q = new Elem(p->Adat,NULL);
            q->next = q;
        }
        last = q;
        }
        current = first;
THIS PART is the error:
[
        while(current!=NULL && current->Adat!=s.current->Adat)
        {
            current=current->next;
        }
]
    }

我刚把它删除了，它还能用。:)

Answer 1

问题是

lista operator+(lista& a, lista& b)
{
    ...
    return a;
}

这将创建a的副本，然后立即销毁(因为您不会将其保存在任何地方)。您没有定义复制构造函数，所以您使用编译器定义的构造函数(它只是复制指针)。临时函数的析构函数会删除a.first所指向的内存，然后当你调用a的析构函数时，一切都会爆炸。

如果析构函数正在释放内存，则需要声明复制构造函数和复制赋值操作符。最简单的定义是删除它们(以= delete;结尾)-然后您需要更改operator +的定义(可能通过将函数名称更改为append())。