Question

I have a linked list, and I want to sort them by names (for example the names "Bx", "Tx", "Ax" would become : "Ax", "Bx", "Tx")... I need to switch the names if the one in the node s right has a "smaller name"..

这是我写的:

typedef struct data
{
char *name;
}data;

typedef struct Node 
{
data NodeData;
struct Node *next;
struct Node *prev;
}Node;

void Sorting(Node *head)
{
 Node *temp = head;
 Node *temp2 = (Node*)malloc(sizeof(Node));
 while (temp != NULL)
 {
       if (1 == (strcmp(temp -> NodeData.name, temp -> next -> NodeData.name))) 
       {
              strcpy (temp2 -> NodeData.name, temp -> NodeData.name);
              strcpy (temp -> NodeData.name, temp -> next -> NodeData.name);
              strcpy (temp -> next -> NodeData.name, temp2 -> NodeData.name);
       } 

       temp = temp -> next;

 }

}

I m getting an runtime - error on the part where I need to swarp betwen the node s name(the strcpy lines): An access violation (segmentation fault)...

Answer 1

if (1 == (strcmp(temp -> NodeData.name, temp -> next -> NodeData.name))) {...}

When this line is executed, there is no guarantee that temp->next is not NULL. If it is NULL, temp->next->NodeData.Name would be rather painfull.

而且,正如已经说过的那样,测试波环(Sstrcmp)结果与1相比是不正确的。扼杀可转折任何价值等于零、零或零。

而且,正如已经说过的那样,扼杀是不正确的;扼杀可能具有不同的分配规模,或者可以生活在不可磨灭的记忆中(固定不变)。浏览点就足够了。

最新资料:

void Sorting(Node *head)
{
 Node *temp ;

                /* since the compare dereferences temp->next you ll have to verify that it is not NULL */
 for (temp = head; temp && temp->next; temp = temp->next)  
 {
       if (strcmp(temp->NodeData.name, temp->next->NodeData.name) > 0 )  
       {
                /* no need for a whole node, since you only copy a pointer */
              char *cp;
              cp = temp->NodeData.name;
              temp->NodeData.name = temp->next->NodeData.name;
              temp->next->NodeData.name = cp;
       } 

 }

}

BTW:划出一个有链接的清单确实令人怀疑。相联的清单比较容易,而且更加精.地被合并。

Answer 2

我不认为你应该通过交换数据价值来对清单进行分类,相反,你应该把标语自己.掉。请注意,这要求你把点子归还名单上新的第一点。

EDIT:如果你重新成为双向点,通过名单负责人的点子也是行之有效的。它可以使法典更加简单。

Answer 3

char* 此处

char *temp2;

但Node 此处

strcpy (temp2 -> NodeData.name, temp -> NodeData.name);

因此,根据这一条,你的行为是:<条码>temp2<>> 代码”是一种结构或结合,但不是。

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