双链表插入分段故障 - C.

时间:2014-02-13 04:44:06

标签: c struct doubly-linked-list

我有一个写作先到先服务和循环罗宾模拟的作业并比较它们。我开始创建一个用于制作事件列表的链表。我已经制作了insert和print_list函数,但是无法使它工作。我的朋友告诉我使用Doubly Linked List,所以我重新处理了链表,并尝试使插入功能正常工作。我解决了大部分问题,但我现在无法找到插入功能的问题。我在While循环语句中得到分段错误,当我比较插入指针数据和当前指针数据时。我已经查看了我的C书,它有与我类似的代码,也通过Doubly Linked List Segmentation Fault -C问题,但仍然无法得到我的错误。

这是我的输出

-bash-3.2$ ./main 3 100
please enter the process-id, arrival-time and cpu-time.
1 2 3 4
something
something
DONE1
DONE2
DONE3
DONE4
Segmentation fault

因为test4(DONE4)打印而下一次测试没有,这意味着我在这部分中有错误

while((currentptr->arrivaltime < newpointer->arrivaltime) && currentptr != NULL)
你能帮帮我吗?我的代码出了什么问题?

struct event {
   struct event *next;
   int processid;
   int arrivaltime;
   int cputime;
};
typedef struct event Event;
typedef Event *eventPtr;

Event create_node(int processid, int arrivaltime, int cputime);
void add_event(Event *newpointer, eventPtr *eventlist);
void print_node(Event node);
void print_eventlist(Event *eventlist);
Event get_nextevent(Event *eventlist);

int main(int argc, char* argv[])
{
int sourcenum,a,b,c,i;
Event tempevent;
eventPtr eventlist = NULL;
print_eventlist(&(*eventlist));
char* sources,timeSlice;
if(argc != 3)
printf("Proper usage is: main.c sources time-slice\n");
else
{
    sourcenum = atoi(argv[1]);
    timeSlice = atoi(argv[2]);
    for(i = 0; i < sourcenum ; i++)
    {
          //print_eventlist(&(*eventlist));
          printf("please enter the process-id, arrival-time and cpu-time.\n");
          scanf("%d %d %d",&a, &b, &c);
          printf("something\n");
          tempevent = create_node(a,b,c);
          print
          printf("something\n");
          add_event(&tempevent, &eventlist);
    }

}
return 0;
}


void add_event(Event *newpointer, eventPtr *eventlist)     //insert
{
 printf("DONE1\n");
 eventPtr currentptr, prevptr;
 printf("DONE2\n");
 prevptr = NULL;
 printf("DONE3\n");
 currentptr = *eventlist;
 printf("DONE4\n");
 while((currentptr->arrivaltime < newpointer->arrivaltime) && currentptr != NULL)                  /*on every loop previous and current pointer gets upgraded*/
 {
         printf("DONEW\n");
         prevptr = currentptr;
         currentptr = currentptr->next;   
 }
 printf("DONEW\n");
 if(prevptr != NULL)
 {
     printf("DONEW\n");
     prevptr->next = newpointer;
     newpointer->next = currentptr;
 }
 else
 {
     printf("DONEW\n");
     newpointer->next = *eventlist;                      //?????
     *eventlist = newpointer;
 }
 printf("DONE\n");
}

2 个答案:

答案 0 :(得分:1)

您说明问题最有可能出现在

while((currentptr->arrivaltime < newpointer->arrivaltime) && currentptr != NULL)

您要测试currentptr!=NULL的事实意味着它可能为NULL。但如果是,那么你病情的早期部分就会失败。首先测试NULL,然后进行比较的后半部分......:

while(currentptr != NULL && (currentptr->arrivaltime < newpointer->arrivaltime))

答案 1 :(得分:1)

eventPtr eventlist = NULL;

add_event(&tempevent, &eventlist);
{
    currentptr = *eventlist;  //currentptr == NULL at this point
    ...............
    while(currentptr->arrivaltime ....)  // Segmentation
}

在您的代码中,“eventlist = NULL”传递到“add_event”。这就是分割的原因。

我认为你应该在“add_event”中使用“malloc()”来获取“eventlist”的条目。