Question

这是一个试图制作链表的程序。

#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;
     }

}

从尚未完成的程序中，我理解这一点：

enter image description here

如果第二次调用函数addNode_AT_END后的数字为真，那么temp2->next语句中的while( temp2->next != NULL )包含什么？

Answer 1

它包含NULL，这是因为这一行：

temp1->next = NULL;

通过执行上述步骤，每个新节点都有next指针，并且新节点会附加在列表的末尾，结果是列表总是NULL。 NULL循环遍历到列表末尾，while设置条件，直到while(temp2->next != NULL) next变为temp2，NULL }}

Answer 2

您的图表不正确。 start = temp2确实表示start和temp2指针都指向同一个节点。您的图表显示temp2指针的下一个字段包含start的地址。在执行start->next = temp1之后，并不意味着如果在temp1中获得一些新节点值（在下一个函数调用中），start->next仍然会指向刚刚分配的新值temp1。它将保留旧的值，然后用新的值覆盖它。 start->next = temp1只是复制temp1中的值即。变量的地址（指针变量）start指向的节点的下一个组件start->next。之后，start和temp1之间没有任何关联。

在链表上下文中“temp1 ----＆gt; temp2”表示地址存储在next中的节点的temp1字段，保存带有地址的节点的地址由temp2举行或举行。现在，在更改指针变量temp2的值后，不会更改next中存储的地址所保存的节点的temp1字段。 temp1->next仍然包含之前存储的值。

下一个链接没有指向某个变量名，也就是说，start->next = temp不会使start节点的下一个节点始终指向temp1包含的任何节点，但是它start->next将包含temp1在分配时存储的地址。

请注意，通过说“start指向temp1”表示地址

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

将在temp2->next = NULL时中断，这意味着temp2指向列表的最后一个节点。此时temp2->next = temp1在temp2当前指向的节点之后链接新分配的节点。这只是在最后添加新节点。

  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.

<强>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

指针是行进/遍历列表的方式。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，那么按顺序跟随下一个字段，我们就可以访问每个节点。 temp1和temp2是遍历完成的指针，它们充当临时指针，temp1用于保存新分配的节点，temp2用于旅行通过列表跟踪next链接直到最后一个，当找到最后一个链接时（由下一个字段中的NULL指针检测到），最后一个节点的NULL链接被新分配的节点替换为temp1。现在，temp1所持有的节点被链接/添加到列表的末尾，temp1被重用来保存另一个新节点。

temp2->接下来包含什么？

2 个答案: