想用c编写一个用来存储多边形顶点的程序，但是结果和预期的不对，可以麻烦帮我看看吗
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悬赏园豆：10
[待解决问题]
#include <stdio.h>
#include <stdlib.h>
//定义六边形结构体
typedef struct hexagon{
int num;
int x;
int y;
struct hexagon* next;
}Hexagon;
//创建新的六边形节点
Hexagon* creat_hexagon(int num,int x,int y){
Hexagon* new_hexagon = (Hexagon*)malloc(sizeof(Hexagon));
new_hexagon->num = num;
new_hexagon->x = x;
new_hexagon->y = y;
new_hexagon->next = NULL;
return new_hexagon;
}
//在链表的结尾添加新的六边形节点
void add_hexagon(Hexagon** head,Hexagon* new_hexagon){
Hexagon* current_hexagon = *head;
if(*head == NULL){
*head = new_hexagon;
}else{
while(current_hexagon->next != NULL){
current_hexagon = current_hexagon->next;
}
current_hexagon->next = new_hexagon->next;
}
}
// 输出链表中存储的六边形信息
void print_hexagon(Hexagon* head) {
while (head != NULL) {
for(int i = 0; i <= 6; i++)
{
printf("六边形的第%d个节点坐标为: (%d,%d)", head->num , head->x, head->y);
head = head + 1;
}
}
}
int main() {
int max = 6;
Hexagon* head = NULL;
int x,y,num;
// 创建并添加新的六边形节点
printf("请输入六边形各点的的序号和坐标：");
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex1 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex1);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex2 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex2);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex3 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex3);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex4 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex4);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex5 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex5);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex6 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex6);
print_hexagon(head);
return 0;
}
所有回答(2)
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我个人觉得你思路有问题
- 六边形有六个点，你应该用数组好一些，这样可以看出来
//定义六边形结构体
typedef struct hexagon{
int num;
int x;//这里用数组int[6]
int y;//这里用数组int[6]
struct hexagon* next;
}Hexagon;
- 按你这个思路，需要用循环输入
每次创建一个六边形，他的num不变，第一个点初始化了，后面还得用同一个num，初始化后面5个点
所以输出六边形的时候，每个节点遍历了六次
for(int i = 0; i <= 6; i++)
{
printf("六边形的第%d个节点坐标为: (%d,%d)", head->num , head->x, head->y);
head = head + 1;
}
int main() {
int max = 6;
Hexagon* head = NULL;
Hexagon* temp = head;//保存链表头节点地址，用于最后从此开始遍历
int x,y,num;
for(num = 0; num<max; i++){
int number = 0;
for(number =0; number< 6； number++){
printf("请输入第 %d 个六边形的第 %d 个坐标：", num, number);
scanf("%d %d", &x, &y);
Hexagon* hex = creat_hexagon(num ,x ,y);//创建新的
add_hexagon(&head, hex1);//添加到head后
head = head.next//head走到最新的节点
}
}
head = temp; // 将链表头结点还给head
print_hexagon(head); // 打印，如果上一行不写，这里可以用temp试试，看能不能遍历
/* 创建并添加新的六边形节点
printf("请输入六边形各点的的序号和坐标：");
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex1 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex1);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex2 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex2);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex3 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex3);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex4 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex4);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex5 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex5);
scanf("%d%d%d", &num, &x, &y);
Hexagon* hex6 = creat_hexagon(num ,x ,y);
add_hexagon(&head, hex6);
print_hexagon(head);
*/
return 0;
}
我没有调试你的代码，我这边没有vc6.0，C编译器
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1、print_hexagon函数里面应该是head = head->next，而不是+1
2、add_hexagon里面最后一条赋值语句应该是current_hexagon->next = new_hexagon;因为要把new_hexagon塞到链表里面
3、print_hexagon函数for(int i = 0; i <= 6; i++)这里只有6条边，应该改成：for(int i = 0; i < 6; i++)或者for(int i = 1; i <= 6; i++)
你用
你的代码
包裹起来，大伙给你说
– echo_lovely 1年前@echo_lovely: 包裹起来是什么意思呀，截张图吗，我是新人不太了解
– 笨小孩罢了 1年前@笨小孩罢了: https://www.cnblogs.com/echo-lovely/p/16215739.html markdown语法，github，博客园，csdn写博客都会用到代码块，那个就是代码块
– echo_lovely 1年前@笨小孩罢了: @笨小孩罢了: 我只是让你把你的代码用代码块包裹下，这样大家可以复制，不用调整缩进
– echo_lovely 1年前@echo_lovely: 好的，谢谢指导，我去看看
– 笨小孩罢了 1年前@echo_lovely你好，下了个软件才发现是调试用的，请问是这样吗？
– 笨小孩罢了 1年前