此代码是为操作系统创建队列 我使用结构来实现我的进程 并使用arr_processes来处理所有这些进程 和new_processes数组根据其到达时间对此进程进行排序
但是当我在visual studio 2010上运行此代码时 它会产生此运行时错误 运行时检查失败＃2 - 变量arr_processes周围的堆栈已损坏！
这是代码
#include <stdio.h>
#include <stdlib.h>
typedef struct
{
int id;
int arr_time;
int serv_time;
int deadline;
} process;
void print_process(process n);
int main()
{
process arr_processes[8];
process new_processes[8];
process real_processes[3];
process ready_processes[5];
process tmp_process[1];
int length_ready;
int i,length,j;
int length_real;
arr_processes[0].id=1;
arr_processes[0].arr_time=12;
arr_processes[0].serv_time=4;
arr_processes[0].deadline=0;
arr_processes[1].id=2;
arr_processes[1].arr_time=10;
arr_processes[1].serv_time=5;
arr_processes[1].deadline=0;
arr_processes[2].id=3;
arr_processes[2].arr_time=9;
arr_processes[2].serv_time=2;
arr_processes[2].deadline=0;
arr_processes[3].id=4;
arr_processes[3].arr_time=8;
arr_processes[3].serv_time=4;
arr_processes[3].deadline=10;
arr_processes[4].id=5;
arr_processes[4].arr_time=5;
arr_processes[4].serv_time=2;
arr_processes[4].deadline=8;
arr_processes[5].id=6;
arr_processes[5].arr_time=3;
arr_processes[5].serv_time=3;
arr_processes[5].deadline=0;
arr_processes[6].id=7;
arr_processes[6].arr_time=2;
arr_processes[6].serv_time=3;
arr_processes[6].deadline=0;
arr_processes[7].id=8;
arr_processes[7].arr_time=1;
arr_processes[7].serv_time=1;
arr_processes[7].deadline=28;
length=sizeof(arr_processes)/sizeof(arr_processes[0]);
printf("\t length of the processes=%i\n\n",length);
printf("\t The Original processes \n\n");
for(i=0;i<8;i++)
print_process(arr_processes[i]);
// now we want to sort the processes according to their arrival time
for(i=0;i<8;i++)
{
new_processes[i]=arr_processes[i];
}
for(i=0;i<length;i++)
{
for(j=0;j<length-i;j++)
{
if((new_processes[j].arr_time)>(new_processes[j+1].arr_time))
{
tmp_process[0]=new_processes[j];
new_processes[j]=new_processes[j+1];
new_processes[j+1]=tmp_process[0];
}
}
}
printf("\t The New processes \n\n");
for(i=0;i<8;i++)
print_process(new_processes[i]); // the new queue
ready_processes[0]=arr_processes[0];
ready_processes[1]=arr_processes[1];
ready_processes[2]=arr_processes[2];
ready_processes[3]=arr_processes[5];
ready_processes[4]=arr_processes[6];
length_ready=sizeof(ready_processes)/sizeof(ready_processes[0]);
// now we want to design the ready queue
for(i=0;i<length_ready;i++)
{
for(j=0;j<length_ready-i;j++)
{
if((ready_processes[j].arr_time)>ready_processes[j+1].arr_time)
{
tmp_process[0]=ready_processes[j];
ready_processes[j]=ready_processes[j+1];
ready_processes[j+1]=tmp_process[0];
}
}
}
printf("\t The ready processes \n\n");
for(i=0;i<length_ready;i++)
print_process(ready_processes[i]); // the ready queue
// now we want to design the ready real queue for the shortest deadline first
// we donnot need to check for the new proesses at each instant of time
//but we need to check for the service time from now
real_processes[0]=arr_processes[3];
real_processes[1]=arr_processes[4];
real_processes[2]=arr_processes[7];
length_real=sizeof(real_processes)/sizeof(real_processes[0]);
for(i=0;i<length_real;i++)
{
for(j=0;j<length_real-i;j++)
{
if((real_processes[j].deadline)>real_processes[j+1].deadline)
{
tmp_process[0]=real_processes[j];
real_processes[j]=real_processes[j+1];
real_processes[j+1]=tmp_process[0];
}
}
}
printf("\t The real processes \n\n");
for(i=0;i<length_real;i++)
print_process(real_processes[i]); // the ready real queue
// removed real process
process removed_real;
removed_real.id=0;
removed_real.arr_time=0;
removed_real.serv_time=0;
removed_real.deadline=0;
process running_process;
running_process.id=0;
running_process.arr_time=0;
running_process.serv_time=0;
running_process.deadline=0;
int counter=0;
int start_time;
while(counter<=28)
{
printf("when time = %i\n\n",counter);
// printf("\t The real processes when the counter=%i \n\n",counter);
// for(i=0;i<length_real;i++)
// print_process(real_processes[i]); // the ready real queue
// first we must check for the real processes
for(i=0;i<length_real;i++)
{
if((counter==real_processes[i].arr_time)
&&((real_processes[i].deadline)-counter)>=(real_processes[i].serv_time))
{
running_process=real_processes[i];
printf("The non zero deadline process is:%i\n",running_process.id);
real_processes[i]=removed_real;
start_time=counter; // real process
while(counter!=(start_time+running_process.serv_time))
{
printf("At time = %i,The Running Process is...\n",counter);
print_process(running_process);
counter++;
}
}
}
counter++;
}
return 0;
}
void print_process(process n)
{
if(n.deadline!=0)
printf("ID=%i\narr_time=%i\nserv_time=%i\ndeadline=%i\n\n\n",n.id,n.arr_time,n.serv_time,n.deadline);
else if(n.deadline==0)
printf("ID=%i\narr_time=%i\nserv_time=%i\n\n\n",n.id,n.arr_time,n.serv_time);
}
答案 0 :(得分：2)
当你用完索引时，这是一个排序示例：
for(i=0; i<length - 1; i++)
{
for(j=i + 1;j<length;j++)
{
if((new_processes[j].arr_time)>(new_processes[i].arr_time))
{
tmp_process[0]=new_processes[j];
new_processes[j]=new_processes[i] ;
new_processes[i]=tmp_process[0] ;
}
}
}
或者，您可以使用标准功能：
void qsort(void *base, size_t nmemb, size_t size,
int (*compar)(const void *, const void *));
定义比较函数：
int compare_by_arr_time(const void* a, const void* b)
{
int a_int = ((const process*)a)->arr_time;
int b_int = ((const process*)b)->arr_time;
return a_int - b_int; // or b_int - a_int
}
使用如下：
qsort(new_processes,
sizeof(new_processes)/sizeof(new_processes[0]),
sizeof(new_processes[0]),
compare_by_arr_time);
答案 1 :(得分：1)
当你走出数组的边界时，你会遇到这些错误。
for(i=0;i<length;i++)
{
for(j=0;j<length-i;j++)
{
if((new_processes[j].arr_time)>(new_processes[j+1].arr_time))
{
tmp_process[0]=new_processes[j];
new_processes[j]=new_processes[j+1] ;
new_processes[j+1]=tmp_process[0] ;
}
}
}
在第一次迭代中，i = 0，j = 0和j必须小于8 - i，即8。
注意表达式j+1。在最外层循环的第一次迭代期间，此表达式将返回[1 ... 9]范围内的值，因此，您将超出new_processes数组的范围。
有你的问题。
编辑：第一个问题后面的for循环中也可能出现此问题。