next 2->next包含什么？

Question

这是一个试图建立链接列表的程序。

#include <iostream>
using namespace std;

struct node {
char name[20];
int age;
int height;
node* next; // Pointer to the next node
 };
 node* startPTR = NULL; // Start Pointer (root)
                   // This pointer permanantly points to the start of the list
                   // Initially there are no nodes

void addNode_AT_END(); // prototype for the function that 'adds nodes at the end'

int main() {
   do {
 addNode_AT_END();
     cout << "Add more ?";
     char ch;
     cin >> ch;
   } while( ch == 'y');
}

void addNode_AT_END() {
node *temp1;
node *temp2;
temp1 = new node;  // We declare space for a pointer item and assign a temporary pointer to it  
                   //*temp1 is the node that it points to
cout << "Enter the name : ";
cin >> temp1->name;
cout << endl << "Enter the age : ";
cin >> temp1->age;
cout << endl << "Enter height : ";
cin >> temp1->height;
temp1->next = NULL; // indicates that this node when inserted in the list will be the last node
  if( startPTR == NULL) {
    startPTR = temp1;  // In the empty list last node will be the first node
  }  else {
        temp2 = startPTR;
        while( temp2->next != NULL )
            temp2 = temp2->next;
        temp2->next = temp1;
     }

}

从这个尚未完成的项目中，我了解到：

​

​

如果第二次调用函数addNode_AT_END后的图为真，那么temp2->next在while( temp2->next != NULL )语句中包含什么？

Answer 1

你的图表不正确。start = temp2确实意味着start和temp2指针都指向同一个节点。您的图表显示了temp2指针的下一个字段保存着开始的地址。而且，在执行start->next = temp1之后，并不意味着如果您在temp1中获得了一些新的节点值(在下一个函数调用中)，start->next仍然会继续指向刚刚在temp1中分配的新值。它将保存旧值，在您用新的值覆盖它之前。start->next = temp1只是在temp1 ie中复制值。指向变量(指针变量)的地址，该变量是由start指向的节点的下一个组件，即start->next。之后，start和temp1之间就没有任何联系了。

在链接列表上下文中，"temp1 -> temp2“是指存储在temp1中的节点的next字段，该节点的地址由temp2持有或持有。现在，在更改指针变量的值之后，temp2不会更改存储在temp1中的地址处节点的next字段。temp1->next仍然包含它以前存储的值。

下一个链接不指向某个变量名称，即，start->next = temp start 将不会使start节点的下一个节点始终指向节点 temp1 包含的任何节点，但是 start->next 将包含在分配时存储的E 236 temp1 <代码>E 140的地址。

请注意，通过说"start is指向temp1“意味着地址

while (temp2->next != NULL)
  temp2 = temp2->next;

当temp2->next = NULL中断时，这意味着temp2指向列表的最后一个节点。此时，temp2->next = temp1将新分配的节点链接到temp2当前指向的节点之后。这就是简单地在末尾添加新节点。

  At the end of the above while loop

                                                              temp2
                                                                |
                                                                V

(start) ----> (n1) ----> (n2) ----> (n3) . . . (n(n-1)) ----> (nn) ----> NULL


   temp2->next = temp1    makes


                                                              temp2
                                                                |
                                                                V

(start) ----> (n1) ----> (n2) ----> (n3) . . . (n(n-1)) ----> (nn) ----> (temp1)--->NULL


 because temp2 holds the address of (nn) therefore linking the new node to the next node of the last node.

更新

第一次：

start = NULL
a new address is allocated and the address stored into temp1 pointer. Also temp->next = NULL
if condition becomes true and temp1 is assigned to start
start = temp1

List state


start = addr1;
 |
 V
(addr1) ----> (NULL)

第二次：

a new node is allocated and the address of the new node is stored into `temp1`. Let this address be `addr2`. Now `temp1` contains the value `addr2`

start is NOT NULL, as start has addr1 in it from the last call.So the else part is true and we get the address of the start node `addr1` into temp2.

temp2 = start;

which means temp2 now points to `addr1`

while loop is encountered. The first iteration the condition `temp2->next != NULL` is FALSE. This is because `temp2` points to `addr1` and the next pointer field of `addr1` has NULL from the last time the function is called. Therefore the while loop terminates.

The next statement does `temp2->next = temp1` . `temp2` points to `addr1` address, and the address of the newly allocated node `addr2` contained in `temp1` is assigned into the next field of the node whose address is stored into `temp2`. Which actually assigns the address `addr2` to the next field of the node identified by the address `addr1`.


temp1 = addr2     after allocation

start = addr1;
 |
 V
(addr1) ----> (NULL)      at begining
 ^
 |
 temp2


after temp2->next = temp1

start = addr1;
 |
 V
(addr1) ----> (addr2) ----> (NULL)      at end
 ^
 |
 temp2

第三次：

temp1 = addr3      new node address allocated

start = addr1;
 |
 V
(addr1) ----> (addr2) ----> (NULL)      at start
 ^
 |
 temp2


start = addr1;
 |
 V
(addr1) ----> (addr2) ----> (NULL)      next iteration after temp2=temp2->next
                 ^                      
                 |
               temp2


we can see temp2->next = NULL and while condition is false. Note that temp2 contains itself the address addr2, and thus temp2->next is NOT addr2, it is NULL.

start = addr1;
 |
 V
(addr1) ----> (addr2) ----> (NULL)      next iteration after temp2=temp2->next
                 ^                      
                 |
               temp2



After linking: temp2->next = temp1;

start = addr1;               temp1         the address addr3 (new node)
 |                             |          is stored in temp1. this address is assigned
 V                             V         to the next node of temp2, replacing NULL
(addr1) ----> (addr2) ----> (addr3) ----> (NULL)      
                 ^                      
                 |
               temp2

指针是在列表中穿行的手段。列表的起始地址保存在指针start中。当每个节点的下一个字段指向下一个节点时，如果我们得到start节点，那么通过依次跟踪下一个字段，我们可以访问每个节点。temp1和temp2是执行遍历的指针，它们充当临时指针，temp1用于保存新分配的节点，而temp2用于通过跟踪next链接遍历列表，直到最后一个链接(在下一个字段中由空指针检测到)时，最后一个节点的空链接被temp1持有的新分配节点替换。由于现在temp1保存的节点被链接/添加到列表的末尾，所以temp1被重用来容纳另一个新节点。

Answer 2

它包含NULL，这是因为这行：

temp1->next = NULL; 

每个新节点都有next指针，通过执行上面的步骤使之成为NULL，并且新节点被追加到列表的末尾，结果，列表的末尾总是NULL。while循环遍历到列表的末尾，while(temp2->next != NULL)设置条件，直到next of temp2变成NULL，do temp2 = temp2->next。