如何在 C 中实现具有多个节点的链表？答案

【问题标题】：How to implement a linked-list with multiple nodes in C?如何在 C 中实现具有多个节点的链表？
【发布时间】：2015-05-10 21:27:49
【问题描述】：

我正在根据用户输入制作链接列表，如下所示：

How Many employees? 4

现在，每个人都有firstnamelastnamerate和zipcode，带有一个链表我正在尝试获取这些输入并根据记录数执行for循环，但我显然我做得不对：

    struct records {
        char first[20];
        char last[20];
        float rate;
        int zip;
        struct node* next;
    };
    void main()
    {
        int i,n;
        printf("Please indicate the number of records : ");
        scanf("%d",&n);
        struct records *head,*conductor;
        head=(struct records*)malloc(n*sizeof(struct records));
        head->next=NULL;
        for (i=0;i<n;i++){
        printf("\nEnter employee information in the format :\nFirstname Lastname rate Zipcode (newline for the next)\n");
        scanf("%s %s %f %d",&head->first,&head->last,&head->rate,&head->zip);
        conductor=head;
        conductor=conductor->next;}
}

我怎样才能做到这一点？

【问题讨论】：

标签： c pointers data-structures linked-list

【解决方案1】：

要修复的样本

struct records {
    char first[20];
    char last[20];
    float rate;
    int zip;
    struct records *next;//typo struct node* next; 
};
int main(void){//return type is `int`
    int i, n;
    printf("Please indicate the number of records : ");
    scanf("%d", &n);

    struct records *head,*conductor;
    head=(struct records*)malloc(n*sizeof(struct records));
    for (i=0; i<n; i++){
        printf("\nEnter employee information in the format :\nFirstname Lastname rate Zipcode (newline for the next)\n");
        scanf("%19s %19s %f %d", head[i].first, head[i].last, &head[i].rate, &head[i].zip);
        head[i].next = &head[i+1];
    }
    head[n-1].next = NULL;

    return 0;
}

【讨论】：