问题

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};
typedef struct ListNode *NodePtr;

int main() {
    NodePtr p, q, r;
    p = (NodePtr)malloc(sizeof(struct ListNode));
    q = (NodePtr)malloc(sizeof(struct ListNode));
    
    p->data = 23;
    p->next = q;
    
    q->data = 45;
    q->next = NULL;
    
    r = (NodePtr)malloc(sizeof(struct ListNode));
    r->data = 34;
    r->next = p->next;
    
    p->next = r;
    
    free(p);
    free(q);
    free(r);
    
    return 0;
}

上述代码创建了一个简单的链表结构，然后添加一个新节点并更改了相应的指针，最后释放了内存。

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};
typedef struct ListNode *NodePtr;

NodePtr p, q, r;
p = (NodePtr)malloc(sizeof(struct ListNode));
q = (NodePtr)malloc(sizeof(struct ListNode));
p->data = 23;
p->link = q;
p->link->data = 45;
q->link = NULL;
r = (NodePtr)malloc(sizeof(struct ListNode));
r->data = 34;
r->link = p->link;
p->link = r;
free(p);
free(q);
free(r);

答案1

得分: 2

以下是翻译好的部分：

1. • code cannot be executed at the global scope, you must enclose the code in a function definition and call it explicitly directly or indirectly from the main() function.
   • 代码不能在全局范围执行，必须将代码封装在一个函数定义中，并直接或间接地从 main() 函数中调用它。
2. • there is no link member in the ListNode structure. You probably mean next instead.
   • ListNode 结构中没有 link 成员。你可能指的是 next。

3. Also note that it is confusing and error prone to hide pointers behind typedefs. Pointers are an important part of the C memory model, making them explicit as struct ListNode *p or ListNode *p improves readability for most programmers.

   • 还请注意，将指针隐藏在 typedef 后面会令人困惑且容易出错。指针是C内存模型的重要组成部分，将它们明确表示为 struct ListNode *p 或 ListNode *p 对大多数程序员来说提高了可读性。

4. Here is a modified version:

   • 以下是修改后的版本：

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};
typedef struct ListNode ListNode;

ListNode *new_node(int data) {
    ListNode *np = malloc(sizeof(*np));
    if (np == NULL) {
        fprintf(stderr, "cannot allocate list node\n");
        exit(1);
    }
    np->data = data;
    np->next = NULL;
    return np;
}

int main(void) {
    ListNode *p = new_node(23);
    ListNode *q = new_node(45);
    ListNode *r = new_node(34);

    r->next = q;
    p->next = r;

    free(p);
    free(q);
    free(r);
    return 0;
}

希望这能帮助你明白原始代码中提到的问题和修改。

#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode *next;
};
typedef struct ListNode ListNode;

ListNode *new_node(int data) {
    ListNode *np = malloc(sizeof(*np));
    if (np == NULL) {
        fprintf(stderr, "cannot allocate list node\n");
        exit(1);
    }
    np->data = data;
    np->next = NULL;
    return np;
}

int main(void) {
    ListNode *p = new_node(23);
    ListNode *q = new_node(45);
    ListNode *r = new_node(34);

    r->next = q;
    p->next = r;

    free(p);
    free(q);
    free(r);
    return 0;
}

答案2

得分: 1

link 是从哪里来的？应该是 next。正如其他人所说，将其写入一个函数/主函数中。

p->link=q;