如何使用c在链表节点中传递两个参数？

I am learning how to use structure with linked list in c. So I was trying to keep inserting the node to first position. Also, each node contains two different kinds of values, one is char array and another one is integer. Once I insert a value, then print the current list out. I look over and over again but I can't find where gets something wrong because I set a printf() test in the insert function. I saw the new_node is what I currently insert.

我正在学习如何在c中使用带有链表的结构。所以我试图继续将节点插入第一个位置。此外，每个节点包含两种不同的值，一种是char数组，另一种是整数。插入一个值后，打印出当前列表。我一遍又一遍地看，但我找不到哪里出错了，因为我在insert函数中设置了printf（）测试。我看到new_node是我目前插入的。

When I insert a 1 first and it prints out a/1. But I am so confusing when I insert another set such as b 2, it prints out

当我先插入1并打印出一个/ 1时。但是当我插入另一组如b 2时，我很困惑，它打印出来

b/2

B / 2

b/1

B / 1

which means the name of first node changed along with inserting a new node, but the numbers did not change. Can anyone see what happened here?

这意味着第一个节点的名称随着插入一个新节点而改变，但数字没有改变。谁能看到这里发生了什么？

struct node{
    char *name;
    int numbers;
    struct node *next;
};
struct node *list = NULL;

void printList(struct node *node){
    while (node != NULL){
        printf("%s/%d\n", node->name, node->numbers);
        node = node->next;
  }
}

void insert(struct node **list, char *name, int numbers){
    struct node * new_node = malloc(sizeof(struct node));
    new_node->name = name;
    new_node->numbers = numbers;
    //printf("......%s %d.........\n", new_node->name, new_node->numbers);
    new_node->next = *list;
    *list = new_node;
}

int main(){
    ......code for scanf() are omitted......
    insert(&list, name, numbers);
    printList(list);
}

2 个解决方案

#1

Your problem is that your are pointing on same buffer for every node. Even without seeing the code you removed, I can guess you are reading the node names in same char array. You would better consider using strdup() as it was suggested:

您的问题是您指向每个节点的相同缓冲区。即使没有看到你删除的代码，我猜你正在读取相同char数组中的节点名称。您最好考虑使用strdup（），因为它建议：

new_node->name = strdup(name);

Make sure you free this pointer when you are destroying the node.

在销毁节点时，请确保释放此指针。

#2

In C, strings take the form of 'character arrays', that is - a series of char variables next to each other, followed by a null character.

在C中，字符串采用'字符数组'的形式，即 - 一系列彼此相邻的char变量，后跟一个空字符。

When you see something like

当你看到类似的东西

char* string = "foo";

in C, that is the declaration of a char*, (a pointer to a variable of type char) assigned the memory address of the first character in a string. In your insert function the line

在C中，这是char *的声明，（指向char类型的变量的指针）分配了字符串中第一个字符的内存地址。在你的插入功能行

new_node->name = name;

takes the memory address assigned to name and stores it in your object. It does not store the string value itself in your object; that remains exactly where it was. If you then use scanf(), passing in the same pointer, you will overwrite the original string, which the object you previously created is still pointing to.

获取分配给name的内存地址并将其存储在对象中。它不会将字符串值本身存储在您的对象中;这仍然是它的确切位置。如果然后使用scanf（），传入相同的指针，您将覆盖原始字符串，您之前创建的对象仍然指向该字符串。

#1

new_node->name = strdup(name);