关于#c语言#的问题：请你完成以下函数代码structNodeconnect(structNodex,structNodey,structNodez){}#include <stdio.h>

建立一个链表，每个结点存储一个学生的姓名和成绩，将a,b,c三个结点按照分数由高到低链接起来，然后输出，编写链表结点连接函数connect。

链表的结构定义方式如下：

struct Node {

char *name;

int  score;

struct Node *next;

};

请你完成以下函数代码

struct Node * connect(struct Node *x, struct Node *y, struct Node *z)
{

}


#include <stdio.h>  
#include <stdlib.h>  
#include <string.h>
  
struct Node {  
    char *name;  
    int  score;  
    struct Node *next;  
};  
  
struct Node *  connect(struct Node *x, struct Node *y, struct Node *z);  
  
int main(int argc, char *argv[]) {      
     struct Node a, b, c, *p;  
       
       
     a.name = "LiPing";  
     b.name = "LiuHai";  
     c.name = "FengYun";   
     scanf("%d,%d,%d", &a.score, &b.score, &c.score);  
     p= connect(&a,&b,&c);  
     printf("%s-%d\n",p->name,p->score);  
     p=p->next;  
     printf("%s-%d\n",p->name,p->score);  
     p=p->next;  
     printf("%s-%d\n",p->name,p->score);  
     return 0;  
}  
struct Node * connect(struct Node *a, struct Node *b, struct Node *c)
{
    struct Node * head,*p,*q,*temp;
    int i=0;
    head=(struct Node *)malloc(sizeof(struct Node));
    head->next=NULL;
    head->next=a;
    a->next=b;
    b->next=c;
    c->next=NULL; 
    q=(struct Node *)malloc(sizeof(struct Node));
    
    temp=a;
    if(b->score>temp->score)
    temp=b;
    if(c->score>temp->score)
    temp=c;
    if(temp!=a)
    {
        q->score=a->score;
        strcpy(q->name,a->name);
        a->score=temp->score;
        strcpy(a->name,temp->name);
        temp->score=q->score;
        strcpy(temp->name,q->name);
    }
    temp=b;
    if(c->score>temp->score)
    temp=c;
    if(temp!=b)
    {
        q->score=b->score;
        strcpy(q->name,b->name);
        b->score=temp->score;
        strcpy(b->name,temp->name);
        temp->score=q->score;
        strcpy(temp->name,q->name);
    }
       return head->next;
}

为什么运行到图中语句时，会报错呢？

题主的代码修改如下，供参考：

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

struct Node {
    char* name;
    int  score;
    struct Node* next;
};

struct Node* connect(struct Node* x, struct Node* y, struct Node* z);

int main(int argc, char* argv[]) {
    struct Node a, b, c, * p;

    a.name = "LiPing";
    b.name = "LiuHai";
    c.name = "FengYun";
    scanf("%d %d %d", &a.score, &b.score, &c.score);
    p = connect(&a, &b, &c);
    printf("%s-%d\n", p->name, p->score);
    p = p->next;
    printf("%s-%d\n", p->name, p->score);
    p = p->next;
    printf("%s-%d\n", p->name, p->score);
    return 0;
}
struct Node* connect(struct Node* a, struct Node* b, struct Node* c)
{
    struct Node* head, * p;    //, * q, * temp;
                               //int i = 0;
    head = (struct Node*)malloc(sizeof(struct Node));
    head->next = NULL;
    head->next = a;
    a->next = NULL;
    if (b->score > a->score) {
        b->next = head->next;
        head->next = b;
    }
    else{
        a->next = b;
        b->next = NULL;
    }
    p = head;
    while (p->next && p->next->score > c->score)
        p = p->next;
    c->next = p->next;
    p->next = c;

#if 0     // 以下删除
    a->next = b;
    b->next = c;
    c->next = NULL;
    q = (struct Node*)malloc(sizeof(struct Node));
    temp = a;
    if (b->score > temp->score)
        temp = b;
    if (c->score > temp->score)
        temp = c;
    if (temp != a)
    {
        q->score = a->score;
        strcpy(q->name, a->name);
        a->score = temp->score;
        strcpy(a->name, temp->name);
        temp->score = q->score;
        strcpy(temp->name, q->name);
    }
    temp = b;
    if (c->score > temp->score)
        temp = c;
    if (temp != b)
    {
        q->score = b->score;
        strcpy(q->name, b->name);
        b->score = temp->score;
        strcpy(b->name, temp->name);
        temp->score = q->score;
        strcpy(temp->name, q->name);
    }
#endif
    return head->next;
}

不知道你这个问题是否已经解决, 如果还没有解决的话:

以下回答来自chatgpt:

由于没有提供具体的出错语句和错误提示，我无法确定导致错误的原因。但是，以下是一些可能导致错误的常见问题和解决方案：

没有正确初始化参数指针。在调用connect函数时，必须传递指向链表头部的指针，否则程序会崩溃。请确保在调用函数之前初始化参数指针，并将它们指向链表头，否则在访问链表节点时会出错。
没有正确连接链表节点。在链接三个节点时，必须正确设置它们之间的指针。请确保按照分数由高到低将节点x，y，z连接起来，即先将x的next指针指向y，再将y的next指针指向z。
没有正确释放内存。在连接链表节点时，可能需要动态分配内存。请确保在使用完节点之后，使用free()函数释放节点的内存。

以下是一个可能的解决方案：

struct Node *connect(struct Node *x, struct Node *y, struct Node *z) {
    struct Node *head, *tail;
    head = tail = NULL;

    // 按照分数由高到低连接节点
    if (x->score >= y->score && x->score >= z->score) {
        head = tail = x;
        if (y->score >= z->score) {
            tail->next = y;
            tail = y;
            tail->next = z;
        } else {
            tail->next = z;
            tail = z;
            tail->next = y;
        }
    } else if (y->score >= x->score && y->score >= z->score) {
        head = tail = y;
        if (x->score >= z->score) {
            tail->next = x;
            tail = x;
            tail->next = z;
        } else {
            tail->next = z;
            tail = z;
            tail->next = x;
        }
    } else {
        head = tail = z;
        if (x->score >= y->score) {
            tail->next = x;
            tail = x;
            tail->next = y;
        } else {
            tail->next = y;
            tail = y;
            tail->next = x;
        }
    }

    // 输出连接结果
    struct Node *ptr = head;
    while (ptr != NULL) {
        printf("%s %d\n", ptr->name, ptr->score);
        ptr = ptr->next;
    }

    // 释放节点内存
    free(x);
    free(y);
    free(z);

    return head;
}

这个解决方案将节点x，y，z按照分数由高到低连接起来，并输出连接后的链表结果。在函数结尾，使用free()函数释放节点的内存，避免程序内存泄漏。

如果你已经解决了该问题, 非常希望你能够分享一下解决方案, 写成博客, 将相关链接放在评论区, 以帮助更多的人 ^-^

struct Node *connect(struct Node *x, struct Node *y, struct Node *z) {
struct Node *head = NULL;
struct Node *temp = NULL;

// 将结点连接成链表
if (x != NULL) {
    head = x;
    temp = x;
    while (temp->next != NULL) {
        temp = temp->next;
    }
}

if (y != NULL) {
    if (head == NULL) {
        head = y;
    } else {
        temp->next = y;
        temp = y;
        while (temp->next != NULL) {
            temp = temp->next;
        }
    }
}

if (z != NULL) {
    if (head == NULL) {
        head = z;
    } else {
        temp->next = z;
        temp = z;
    }
}

// 对链表进行降序排序
struct Node *curr, *nextNode;
int swapped;

do {
    swapped = 0;
    curr = head;
    
    while (curr->next != NULL) {
        nextNode = curr->next;
        
        if (curr->score < nextNode->score) {
            // 交换节点位置
            struct Node *tempName = curr->name;
            int tempScore = curr->score;
            
            curr->name = nextNode->name;
            curr->score = nextNode->score;
            
            nextNode->name = tempName;
            nextNode->score = tempScore;
            
            swapped = 1;
        }
        
        curr = curr->next;
    }
} while (swapped);

return head;
}

关于#c语言#的问题：请你完成以下函数代码structNode*connect(structNode*x,structNode*y,structNode*z){}#include <stdio.h>

关于#c语言#的问题：请你完成以下函数代码structNodeconnect(structNodex,structNodey,structNodez){}#include <stdio.h>