这里使用指针创造链表，输出为什么不是“1 2 3”，而是“1 10 10”？

Question

这里使用指针创造链表，输出为什么不是“1 2 3”，而是“1 10 10”？

Answer 1

这么改下，看得更清楚点，供参考：

#include <stdio.h>
#include <stdlib.h>
struct Linkedlist{
    int  value;
    struct Linkedlist* next;
};
int main()
{
    int num[10] = {1,2,3,4,5,6,7,8,9,10};

    struct Linkedlist head0 = {num[0],NULL};  //定义第一个结点
    struct Linkedlist *temp= &head0;   //定义结构体指针temp ，指向第一个结点

    struct Linkedlist head1 = {num[1],NULL}; //定义第二个结点
    temp->next = &head1;

    temp = temp->next;

    struct Linkedlist head2 = {num[2],NULL};  //定义第三个结点
    temp->next = &head2;

    temp = temp->next;

    struct Linkedlist head3 = {num[3],NULL};  //定义第四个结点
    temp->next = &head3;

    temp = temp->next;

    struct Linkedlist head4 = {num[4],NULL};
    temp->next = &head4;

    temp = temp->next;

    struct Linkedlist head5 = {num[5],NULL};
    temp->next = &head5;

    temp = temp->next;

    struct Linkedlist head6 = {num[6],NULL};
    temp->next = &head6;

    temp = temp->next;

    struct Linkedlist head7 = {num[7],NULL};
    temp->next = &head7;

    temp = temp->next;

    struct Linkedlist head8 = {num[8],NULL};
    temp->next = &head8;

    temp = temp->next;

    struct Linkedlist head9 = {num[9],NULL};
    temp->next = &head9;

    for(temp = &head0;temp != NULL;temp = temp->next)
        printf("%d\t",temp->value);

    return 0;
}