删除函数正在使程序崩溃 c++

Question

/********************************************
*   Remove the last employee from the list  *
********************************************/
void EmployeeList::Remove() 
{
    newEmployee * nextToEnd = head, 
                * last = head->Next();

    //THIS IS THE PROBLEM
    //no nodes
    if(head == NULL)
        return;

    //remove the only employee in the list
    if(head->Next()== NULL)
    {
        cout << "\n\t\tEmployee ID " << head->empID() << " and salary $" 
             << head->ySalary() << " have been removed.\n"; 
        head = NULL;
        delete head;
    }

    else
    {  
       // remove the last employee of the list
        while(last->Next() != NULL)
        {
            nextToEnd = last;
            last = last->Next();
        }   

       cout << "\n\t\tEmployee ID " << last->empID() << " and salary $" 
            << last->ySalary() << " have been removed.\n";  

        delete last;
        nextToEnd->SetNext(NULL);      
    }
}

尝试从空列表中删除时遇到问题。我知道如果它是空的，我无法删除，但我不会让程序崩溃以显示“员工列表为空。

我指定了我认为问题所在，希望有人能帮我解决。

Answer 1

您所要做的就是代码中已经显示的内容。注意它是如何使用cout 将文本输出到控制台的。更改您指定“问题”的if 语句是输出一条消息然后返回。

但是您的程序崩溃与您标记的内容无关。它因为head->Next() 而崩溃。您不能在 NULL 的对象上调用方法。这应该发生在if (head == NULL) 检查之后。