C语言程序不能成功运行
两个关于疫情数据的代码第一个是用数组完成的，第二个是用链表完成的，两个都不能成功运行数组的可以打开文件但不能进入循环，链表的运行之后无任何结果，且都没有报错和警告，调试的报错只有这个，请大家看看是怎么回事
#include
#define population 3000000
#define cost_of_lock_down_population 60000000
#define ratio 3
#define shelter_size 1000
#define shelter_time 7
#define cost_of_shelter 200
int day[40] = { 1 };
int infection[40] = { 1 };
int shelter = 1;
long cost[40] = { 200 };
int ID[population] = { 1 };
int new[40] = { 1 };
long sum = 0;
int t[40] = { 1 };
int shelter_[cost_of_lock_down_population / shelter_size] = { 1 };
int main()
{
FILE* fp1 = fopen("infected_day_x.txt", "a+");
FILE* fp2 = fopen("cost_of_lock_down_population_day_x.txt", "a+");
FILE* fp3 = fopen("shelter_day_x.txt", "a+");
for (int i = 1; i <= 47; i++)
{
fprintf(fp1, "\n第%d天\t感染人数:%d\n感染的ID为:\t", day[i - 1], infection[i - 1]);
fprintf(fp2, "第%d天\t方舱消耗金额:%ld\n", day[i-1],cost[i-1]);
day[i] = day[i - 1] + 1;
while (i < 40)
{
new[i] = new[i - 1] * ratio;
infection[i] += new[i];
for (int b = 0; b < new[i]; b++)
{
ID[t[i - 1] + b] = 1;
}
t[i] = new[i] + t[i - 1];
}
while (day[i] > shelter_time)
{
for (int b = 1; b <= new[i - 7]; b++)
{
ID[t[i - 7] - b] = 0;
}
infection[i] -= new[i - 7];
if (infection[i] <= 0)
{
break;
}
}
for (int b = 0; b < t[i]; b++)
{
if (ID[b] != 0)
{
fprintf(fp1, "%d\t", b+1);
int n = 0;
n++;
if (n <= shelter_size)
{
fprintf(fp3, "\n方舱%d\nID为%d\t", shelter, b + 1);
}
else
{
shelter++;
n -= shelter_size;
}
shelter_[i] = shelter;
}
}
cost[i] = infection[i] * cost_of_shelter;
sum += cost[i];
sum += cost_of_lock_down_population;
}
fclose(fp1);
fclose(fp2);
fclose(fp3);
FILE* fp4 = fopen("total_cost.txt", "a+");
fprintf(fp4, "总消耗数为%ld", sum);
fclose(fp4);
FILE* fp5 = fopen("infected_day.csv", "a+");
FILE* fp6 = fopen("cost_day.csv", "a+");
FILE* fp7 = fopen("shelter_day.csv", "a+");
for (int i=0; i<47;i++)
{
fprintf(fp5, "%d",day[i]);
fprintf(fp5, "\n");
fprintf(fp5, "%d\n",infection[i]);
fprintf(fp6, "%d",day[i]);
fprintf(fp6, "\n");
fprintf(fp6, "%d\n",cost[i]);
fprintf(fp7, "%d",day[i]);
fprintf(fp7, "\n");
fprintf(fp7, "%d\n",shelter_[i]);
}
fclose(fp5);
fclose(fp6);
fclose(fp7);
return 0;
}
```c
#include
#include
#define population 3000000
#define cost_of_lock_down_population 60000000
#define ratio 3
#define shelter_size 1000
#define shelter_time 7
#define cost_of_shelter 200
typedef struct _id
{
int number;
struct _id* next2;
}id;
typedef struct _ID
{
int a;
id* ids;
struct _ID* next3;
}ID;
typedef struct _num
{
int day;
int infection;
long cost;
int shelter;
}num;
typedef struct _node
{
num n;
ID* IDs;
struct _node* next;
}node;
int main()
{
node* head = NULL;
id* head2 = NULL;
ID* head3 = NULL;
int i = 1;
int b = 0;
long sum = 0;
int new = 0;
for (; i <= 40; i++)//从开始到结束共计40天
{
node* p = (node*)malloc(sizeof(node));
if (p == NULL)
{
printf("no");
exit(0);
}
if (i == 1)//赋初值
{
p->n.day = 1;
p->n.infection = 1;
p->n.cost = 0;
p->n.shelter = 1;
new = (p->n.infection) * ratio;
}
else
{
(p->n.day)++;
for (int t = p->n.infection + 1; t <= new * ratio; t++)
{
id* p2 = (id*)malloc(sizeof(id));
if (p2 == NULL)
{
printf("no");
exit(0);
}
p2->number = t;
p2->next2 = NULL;
id* last2 = head2;
if (last2)
{
while (last2->next2)
{
last2 = last2->next2;
}
last2->next2 = p2;
}
else
{
head2 = p2;
}
b++;
}
ID* p3 = (ID*)malloc(sizeof(ID));
if (p3 == NULL)
{
printf("no");
exit(0);
}
p3->a = b;
p3->ids = head2;
p3->next3 = NULL;
ID* last3 = head3;
if (last3)
{
while (last3->next3)
{
last3 = last3->next3;
}
}
else
{
head3 = p3;
}
while (i <= 15)
{
new *= ratio;
}
p->n.infection += new;
while (p->n.day > shelter_time)
{
ID* phead3 = head3->next3;
p->n.infection -= p3->a;
free(head3);
head3 = phead3;
}
if (p->n.infection <= 0)
{
break;
}
p->n.cost = (p->n.infection) * cost_of_shelter;
p->n.shelter = (p->n.infection) % shelter_size + 1;
p->IDs = head3;
p->next = NULL;
node* last = head;
if (last)
{
while (last->next)
{
last = last->next;
}
last->next = p;
}
else
{
head = p;
}
}
sum += p->n.cost;
sum += cost_of_lock_down_population;
}
FILE* fp1 = fopen("infected_day_x.txt", "a+");
node* move = head;
id* move2 = move->IDs->ids;
ID* move3 = move->IDs;
while (move != NULL)
{
fprintf(fp1, "\n第%d天\t感染人数:%d\n感染的ID为:\t", move->n.day, move->n.infection);
while (move3 != NULL)
{
while (move2 != NULL)
{
fprintf(fp1, "%d\t", move2->number);
move2 = move2->next2;
}
move3 = move3->next3;
}
move = move->next;
}
fclose(fp1);
FILE* fp2 = fopen("cost_of_lock_down_population_day_x.txt", "a+");
while (move != NULL)
{
fprintf(fp2, "第%d天\t方舱消耗金额:%ld\n", move->n.day, move->n.cost);
move = move->next;
}
fclose(fp2);
FILE* fp3 = fopen("shelter_day_x.txt", "a+");
while (move != NULL)
{
fprintf(fp3, "第%d天\t方舱使用数量为%d\n", move->n.day, move->n.shelter);
move = move->next;
}
fclose(fp3);
FILE* fp4 = fopen("total_cost.txt", "a+");
fprintf(fp4, "总消耗数为%ld", sum);
fclose(fp4);
FILE* fp5 = fopen("infected_day.csv", "a+");
FILE* fp6 = fopen("cost_day.csv", "a+");
FILE* fp7 = fopen("shelter_day.csv", "a+");
while (move != NULL)
{
fprintf(fp5, "%d", move->n.day);
fprintf(fp5, "\n");
fprintf(fp5, "%d\n", move->n.infection);
fprintf(fp6, "%d", move->n.day);
fprintf(fp6, "\n");
fprintf(fp6, "%d\n", move->n.cost);
fprintf(fp7, "%d", move->n.day);
fprintf(fp7, "\n");
fprintf(fp7, "%d\n", move->n.shelter);
move = move->next;
}
fclose(fp5);
fclose(fp6);
fclose(fp7);
return 0;
}
```
你的代码中有一个问题是在这个循环中：
for (int i = 1; i <= 47; i++)
你在循环中又嵌套了一个循环：
while (i < 40)
而这个嵌套循环的判断条件与外层循环的判断条件相矛盾，所以会导致程序死循环。
你可以把这个嵌套循环改为：
for (int j = 1; j <= 40; j++)
另外，你的程序会计算47天的疫情数据，而在前40天的循环中，你的文件输出和变量计算都是以40天为标准的，所以在第41天到第47天的输出会有问题。
最后，在输出"总消耗数"时，你的程序会计算47天的方舱费用，但是每天的方舱费用是没有累加的，所以这个费用数字会过大。
#include<stdio.h>
#define population 3000000
#define cost_of_lock_down_population 60000000
#define ratio 3
#define shelter_size 1000
#define shelter_time 7
#define cost_of_shelter 200
int day[47] = { 1 };
int infection[47] = { 1 };
int shelter = 1;
long cost[47] = { 200 };
int ID[population] = { 1 };
int new[47] = { 1 };
long sum = 0;
int t[47] = { 1 };
int shelter_[cost_of_lock_down_population / shelter_size] = { 1 };
int main()
{
FILE* fp1 = fopen("infected_day.txt", "a+");
FILE* fp2 = fopen("cost_of_lock_down_population_day.txt", "a+");
FILE* fp3 = fopen("shelter_day.txt", "a+");
for (int i = 1; i <= 47; i++)
{
fprintf(fp1, "\n第%d天\t感染人数:%d\n感染的ID为:\t", day[i - 1], infection[i - 1]);
fprintf(fp2, "第%d天\t方舱消耗金额:%ld\n", day[i-1],cost[i-1]);
day[i] = day[i - 1] + 1;
for (int j = 1; j <= 40; j++)
{
new[i] = new[i - 1] * ratio;
infection[i] += new[i];
for (int b = 0; b < new[i]; b++)
{
ID[t[i - 1] + b] = 1;
}
t[i] = new[i] + t[i - 1];
}
while (day[i] > shelter_time)
{
for (int b = 1; b <= new[i - 7]; b++)
{
ID[t[i - 7] - b] = 0;
}
infection[i] -= new[i - 7];
if (infection[i] <= 0)
{
break;
}
}
for (int b = 0; b < t[i]; b++)
{
if (ID[b] != 0)
{
fprintf(fp1, "%d\t", b+1);
int n = 0;
n++;
if (n <= shelter_size)
{
fprintf(fp3, "\n方舱%d\nID为%d\t", shelter, b + 1);
}
else
{
shelter++;
n -= shelter_size;
}
shelter_[i] = shelter;
}
我帮你改了部分代码，你试试
第28行~37行的那个while循环是死循环。while改成if后，数组实现的貌似没有bug了。
程序效率很低，但是不知道你要实现什么功能，不知道怎么改，你把功能说一下吧。
这个问题可能是因为您的循环条件有误，导致程序无法进入循环。例如，在第一个循环中，您使用了i < 40作为循环条件，但是在循环体内部又使用了i来计算新的感染人数。这可能导致程序无法进入循环。
另外，在链表部分，您在循环中没有进行任何链表操作，也没有定义链表变量，所以无法进行链表操作。
建议重新检查您的循环条件，并确保程序能够正确进入循环，并添加链表操作。
这仔细看了眼你贴的代码，发现题主你的代码行循环存在循环指针溢出/越界的问题，你可以着重检查下你那两个循环，并且你这些嵌套执行效果低下，你也可以贴下 你要实现的需求，或者我可以给你提供思路或者代码实现都行。
你贴出来的代码23行处的for循环的循环条件导致数组访问越界
for (int i = 1; i <= 47; i++)
应为
for (int i = 1; i <= 40 i++)
这个错误信息表明程序试图访问了不能访问的内存，这可能是因为程序中存在越界访问、野指针等问题。在这种情况下，需要检查代码中的数组、指针等变量的访问是否正确，并修复错误。另外，还需要注意编译选项和编译环境的影响，如果编译选项不当会导致编译出的程序存在隐藏问题。
- 看下这篇博客，也许你就懂了，链接：vscode运行c语言 程序
检查文件路径是不是对的
检测下指针或循环越界问题吧
用的什么工具呢? 打断点调试下, 在异常的代码前一行打个断点, 看下变量值, 一般就能找到异常的原因
这个错误是数组超出了范围，指针越界了哇