打印函数链表 C++
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Print Function Linked Lists C++
提问by Tóth Attila
I'm currently learning Linked Lists in C++, and I can't write a print function which prints out the elements of the list; I mean I wrote the function but it doesn't work properly.
我目前正在学习 C++ 中的链表，我无法编写打印列表元素的打印函数；我的意思是我写了这个函数，但它不能正常工作。
#include <iostream>
using namespace std;
struct node
{
char name[20];
node* next;
};
node* addNewPerson(node* head)
{
node* person = new node;
cout << "Name: ";
cin >> person->name;
person->next = NULL;
if (head == NULL) //is empty
{
head = person;
}
else
{
person = person->next;
}
return head;
}
void printList(node* head)
{
node* temp = head;
cout << temp->name << endl;
}
int main()
{
node* head = NULL;
node* temp = head;
unsigned short people = 0;
cout << "How many people do you want to invite to the party?" << endl;
cout << "Answer: ";
cin >> people;
cout << endl;
for (unsigned short i = 1; i <= people; i++)
{
addNewPerson(head);
cout << endl;
}
cout << "LIST: " << endl;
cout << endl;
while (temp != NULL)
{
printList(temp);
temp = temp->next;
}
cin.get();
}
I am wondering what I'm doing wrong, please help me!
我想知道我做错了什么，请帮助我！
回答by Vlad from Moscow
It is obvious that function addNewPerson is wrong.
很明显，函数 addNewPerson 是错误的。
node* addNewPerson(node* head)
{
node* person = new node;
cout << "Name: ";
cin >> person->name;
person->next = NULL;
if (head == NULL) //is empty
{
head = person;
}
else
{
person = person->next;
}
return head;
}
You allocated new node person.
您分配了新的节点人员。
node* person = new node;
Set its field next to NULL
将其字段设置为 NULL
person->next = NULL;
Then if head is not equal to NULL you set person to person->next
然后，如果 head 不等于 NULL，则将 person 设置为 person->next
person = person->next;
As person->next was set to NULL it means that now also person will be equal to NULL.
由于 person->next 被设置为 NULL，这意味着现在 person 也将等于 NULL。
Moreover the function returns head that can be changed in the function. However you ignore returned value in main
此外，该函数返回可以在函数中更改的 head。但是你忽略了 main 中的返回值
addNewPerson(head);
At least there should be
至少应该有
head = addNewPerson(head);
The valid function addNewPerson could look the following way
有效函数 addNewPerson 可能如下所示
node* addNewPerson(node* head)
{
node* person = new node;
cout << "Name: ";
cin >> person->name;
person->next = head;
head = person;
return head;
}
And in main you have to write
主要你必须写
for (unsigned short i = 1; i <= people; i++)
{
head = addNewPerson(head);
cout << endl;
}
Function printList does not output the whole list. It outputs only data member name of the first node that is of head.
函数 printList 不输出整个列表。它只输出首个节点的数据成员名称。
void printList(node* head)
{
node* temp = head;
cout << temp->name << endl;
}
It should look the following way
它应该如下所示
void printList(node* head)
{
for ( ; head; head = head->next )
{
cout << head->name << endl;
}
}
And at last main should look as
最后主要应该看起来像
int main()
{
node* head = NULL;
unsigned short people = 0;
cout << "How many people do you want to invite to the party?" << endl;
cout << "Answer: ";
cin >> people;
cout << endl;
for ( unsigned short i = 0; i < people; i++ )
{
head = addNewPerson(head);
cout << endl;
}
cout << "LIST: " << endl;
cout << endl;
printList( head );
cin.get();
}
回答by user1052842
In your addNewPerson function person->next never gets set, because head is always NULL.
在您的 addNewPerson 函数中，person->next 永远不会被设置，因为 head 始终为 NULL。
回答by Alex Chen
When you are adding a new person, it is not actually put into the linked list, since you have never invoked something like head -> next = person.
当您添加一个新人时，它实际上并未放入链表中，因为您从未调用过类似head -> next = person.
The section in your addNewPersonroutine should be something like
你的addNewPerson例程中的部分应该是这样的
if (head == NULL) {
head = person;
} else {
node * end = head;
while (end -> next != NULL) // find the end of the list to insert new person
end = end -> next;
end -> next = person;
}
And in the beginning you can't just pass headinto addNewPerson, since it is a pointer with null value, the addNewPersonmethod accepts a pointer by value (value of the memory address it is pointing at) rather than a reference to head. So you need to use
而在开始的时候你不能只是通过head进入addNewPerson，因为它是空值的指针，该addNewPerson方法接受由值指针（它在指向的内存地址的值），而不是一个参考head。所以你需要使用
node* addNewPerson(node* &head) { ... }
instead. Finally, when you declare temp, it first receives the value of headas well, so later access to tempwould only give you NULLagain. You need to change it to
反而。最后，当您声明 时temp，它首先接收到headas的值，因此以后访问temp只会NULL再次给您。你需要把它改成
node* &temp = head;
回答by Jonny Henly
addNewPerson method
addNewPerson 方法
Your addNewPersonmethod never added a new person to the list it just set the head to the new person.
您的addNewPerson方法从未将新人添加到列表中，它只是将头部设置为新人。
node* addNewPerson(node* head)
{
node* person = new node;
cout << "Name: ";
cin >> person->name;
person->next = NULL;
if (head->next == NULL) //is empty
{
head->next = person;
}
else
{
person->next = head->next;
head->next = person;
}
return head;
}
printList method
打印列表方法
Your method printListshould do what it says and print the list, instead of just printing one person at a time.
你的方法printList应该按照它说的做并打印列表，而不是一次只打印一个人。
void printList(node* head)
{
node* tmp = head;
while(tmp->next != NULL) {
tmp = tmp->next;
cout >> tmp->name >> endl;
}
}
main method
主要方法
Upadted your main method with the new printListmethod.
使用新printList方法更新了您的主要方法。
int main()
{
node* head = new node;
unsigned short people = 0;
cout << "How many people do you want to invite to the party?" << endl;
cout << "Answer: ";
cin >> people;
cout << endl;
for (unsigned short i = 1; i <= people; i++)
{
addNewPerson(head);
cout << endl;
}
cout << "LIST: " << endl;
cout << endl;
cout << printList(head);
cin.get();
}
Also why are you using unsigned shortand chararrays? Do you have a limited amount of memory? Why not just use intand string?
另外你为什么使用unsigned short和char数组？你的内存有限吗？为什么不直接使用intand string？
回答by user3307862
I think the problem is in how the link list is being added to in addNewPerson.
我认为问题在于如何在 addNewPerson 中添加链接列表。
The input to addNewPerson is 'head' so each time, before adding a new node to the linked list, we need to traverse all the way down the linked list till the last node before appending 'person' to the linked list.
addNewPerson 的输入是'head'，所以每次在向链表添加新节点之前，我们需要一直向下遍历链表直到最后一个节点，然后将'person'附加到链表。
回答by S421
struct node
{
char name[20];
node* next;
};
node* addNewPerson(node* head)
{
node* person = new node;
//cout << "Name: ";
cin >> person->name;
person->next = head;
head = person;
return head;
}
void PrintLL(node *head)
{
while (head!=NULL){
cout<< head->name <<endl;
head=head->next;
}
}
int main() {
node* head = NULL;
node* temp = head;
int num_ppl;
cout << "How many people do you want to invite to the party?" << endl;
cout << "Answer: ";
cin >> num_ppl;
for(int arr_i = 0; arr_i < num_ppl; arr_i++){
head = addNewPerson(head);
}
PrintLL(head);
return 0;
}
