[英]Linked list of pointers C++
I have a list, but now I have to link it. 我有一个列表,但是现在我必须链接它。
Here is my program ( I deleted code inside functions, to make my program more easy to read ). 这是我的程序(为了使程序更易于阅读,我删除了函数内部的代码)。
#include <iostream>
using namespace std;
struct Student
{
char ime[16];
char priimek[16];
char vpisna[10];
char ocenaRV[10];
char ocenaDN[10];
char ocenaKV[10];
char ocenaVI[10];
Student *next;
};
void clean(Student* pointer,int x) // Delete random data
{
}
void dodajanje(int x,Student* s) // Add information about student
{
}
void brisi(Student* pointer,int x) // Delete information about student
{
}
int main()
{
int student,mesto, brisanje, ali = 0;
cout << "Number of students?." << endl;
cin >> student;
Student* s = new Student[student];
clean(s,student);
cout << endl;
cout << "Add student to i place in array." << endl;
cin >> mesto;
dodajanje( mesto, s );
for(int i=0;i<(student*2);i++)
{
cout << "add student = 1, delete student = 2, cout information = 3"<<endl;
cin>>ali;
if (ali == 1)
{
cout << endl;
cout << "Add student to i place in array." << endl;
cin >> mesto;
dodajanje( mesto, s );
}
else if (ali == 2)
{
cout << "delete student on i place ?" << endl;
cin >> brisanje;
brisi(s,brisanje);
}
else
{
break;
}
}
delete[] s;
return 0;
}
Can someone explain me how to link my list, because code in all tutorials I came across was similar to this: 有人可以解释一下如何链接我的列表,因为我遇到的所有教程中的代码都与此类似:
Node* temp = Node();
But in my program my code is: 但是在我的程序中,我的代码是:
Student* s = new Student[student];
And now I'm lost; 现在我迷路了;
Note: I have to create dynamically linked list. 注意:我必须创建动态链接列表。
Linked list is a node based data structure. 链表是基于节点的数据结构。 What you are trying to do is creating a dynamic array of students not a linked list.
您尝试做的是创建一个动态的学生数组,而不是一个链表。
If you really need to create a linked list, in place of following line 如果您确实需要创建链接列表,请代替以下行
Student* s = new Student[student];
you should create a nodes as follows in number of time of students in a for loop and link each other by updating student()-> next= next_student (Psuedo code)
您应在for循环中按以下方式创建一个节点,以增加学生的数量,并通过更新
student()-> next= next_student (Psuedo code)
Student* s = new Student;
And at end, you have to call delete s
within a for loop to deallocate the memory. 最后,您必须在for循环中调用
delete s
来释放内存。
Node* temp = Node();
Node * temp = Node();
This creates a single Node
instance. 这将创建一个
Node
实例。 Though it should be this instead: 虽然应该是这样:
Node* temp = new Node;
Student* s = new Student[student];
学生* s =新学生[学生];
This creates an array of student
number of Student
instances. 这将创建数组
student
数量Student
的情况。 This defeats the purpose of a linked list , as you won't be able to add/remove Student
instances from the array efficiently. 这违反了链接列表的目的,因为您将无法有效地从数组中添加/删除
Student
实例。 But, just for the sake of argument, lets say you really need an array. 但是,仅出于争论的目的,可以说您确实需要一个数组。 You can "link" the
Student
instances together like this: 您可以像这样将“
Student
实例“链接”在一起:
for (int i = 0; i < (student-1); i++)
s[i].next = &s[i+1];
s[student-1].next = NULL;
If you actually need a linked list then you need something more like this instead: 如果您实际上需要一个链表,那么您需要的是这样的东西:
Student *studentList = NULL;
Student *lastStudent = NULL;
for (int i = 0; i < student; ++i)
{
Student* s = new Student;
s->next = NULL;
if (lastStudent) lastStudent->next = s;
if (!studentList) studentList = s;
lastStudent = s;
}
// use studentList as needed...
Student *s = studentList;
while (s)
{
Student *next = s->next;
delete s;
s = next;
}
After fixing that, consider using the STL std::list
class instead, or even std::forward_list
in C++11. 解决此问题后,请考虑改用STL
std::list
类,甚至在C ++ 11中使用std::forward_list
。
That being said, you also need to rethink your code design. 话虽如此,您还需要重新考虑代码设计。 A linked list grows and shrinks dynamically, so there is no point in asking the user for the number of students up front, or pre-allocating the list with garbage that has to be cleaned before it can be used.
链接列表会动态增长和收缩,因此毫无疑问地要求用户预先提供学生人数,或者为列表预先分配必须清除的垃圾才能使用。 Change your loop to run forever (or at least until the user says to stop).
将循环更改为永久运行(或至少直到用户说要停止运行)。 On each iteration, ask the user what to do.
在每次迭代中,询问用户该怎么做。 If
Add
, add a new Student
to the list at that time. 如果
Add
,则此时将一个新的Student
添加到列表中。 If Delete
, ask the user which student to delete, find that Student
, unlink it, and delete
it. 如果为
Delete
,则询问用户要删除的学生,找到该Student
,取消链接,然后delete
其delete
。 If Display
, ask the user which student to display, find that Student
, and display it. 如果是
Display
,则询问用户要显示哪个学生,找到该Student
并显示它。 And so on. 等等。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.