代码为什么加break才能输出结果？

Question

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define LA 4
#define LB 5
struct student{
    int num;
char name[8];
struct student * next ;
}
a[LA],b[LB];
int main() {
    struct student *create(struct student *, int);
    struct student *del(struct student*, struct student*);
    void print(struct student*);
    struct student a[LA]={{101,"Wang"},{102,"Li"},{105,"Zhang"}, {106,"Wei"}};
    struct student b[LB]={{103,"Zhang"},{104,"Ma"}, {105,"Chen"},{107,"Guo"},{108,"1ui"}};
    struct student *headA= create(a, LA), *headB = create(b, LB), *p;
    p = del(headA, headB);
    print(p);
    return 0;
}

//创建
struct student *create(struct student *h, int n) {
    struct student *p, *s, *head;
    head = NULL;
    int i = 0;
    s = (struct student*) malloc(sizeof(struct student));
    p = (struct student*) malloc(sizeof(struct student));
    p->next = NULL;
    head = p;
//    尾插法
    do{
        s->num = h[i].num;
        strcpy(s->name, h[i].name);
        p->next = s;
        p = s;
        s = (struct student*) malloc(sizeof(struct student));
        i++;
    } while (i<n);
    p->next = NULL;
    return (head);
}

//删除
struct student *del(struct student* a, struct student* b){
    struct student *p = a->next, *q = b->next, *pre;
    while (q != NULL) {
        if( p == NULL) {
            p = a->next;
            q = q->next;
        }
        if(p->num == q->num) {
            pre->next = p->next;
            p = pre->next;
            q = q->next;
            break;//求问：为什么这里要break出去才能获得数据？
        }
        pre = p;
        p = p->next;
    }
    return (a);
}

//输出
void print(struct student* head) {
    struct student *p = head->next;
    while (p!=NULL) {
        printf("%d ", p->num);
        p = p->next;
    }
}

就是删除函数中

if(p->num == q->num)

语句，要加break才能得到数据，否则就输出为空，不知道是什么原因？

Answer 1

六堰飞鱼 上午好☀️☀️☀️️
本答案参考ChatGPT-3.5

原因： 在删除函数中，当找到匹配的节点时，需要删除该节点并继续向后查找其他匹配的节点。但是如果没有加上break语句，会导致循环继续执行，继续查找匹配的节点，此时pre指向的是已删除节点的前一个节点，而p指向的是已删除节点的下一个节点，因此会跳过被删除的节点，导致最终输出结果为空。

解决方案：

1. 在找到匹配节点后，加上break语句，跳出循环，继续执行后续代码。

修改后的代码如下：

//删除
struct student *del(struct student* a, struct student* b){
    struct student *p = a->next, *q = b->next, *pre;
    while (q != NULL) {
        if( p == NULL) {
            p = a->next;
            q = q->next;
        }
        if(p->num == q->num) {
            pre->next = p->next;
            p = pre->next;
            q = q->next;
            **break;**
        }
        pre = p;
        p = p->next;
    }
    return (a);
}

1. 不使用break语句，而是在删除节点后将p和q分别向后移动一位，继续查找匹配的节点。

修改后的代码如下：

//删除
struct student *del(struct student* a, struct student* b){
    struct student *p = a->next, *q = b->next, *pre;
    while (q != NULL) {
        if( p == NULL) {
            p = a->next;
            q = q->next;
        }
        if(p->num == q->num) {
            pre->next = p->next;
            p = p->next;
            q = q->next;
        }
        else {
            pre = p;
            p = p->next;
        }
    }
    return (a);
}

以上两种方案都可以正确输出结果。