我想删除所有具有相同idteam作为key的节点,但它会崩溃…我知道它也应该释放()内存,但无论如何我认为这应该工作:S
//defining the struct
struct players {
int idplayer;
int idteam;
struct players *next;
};
struct players *first, *last;
//the function to delete the nodes
void delete(int key){
struct players *post;
struct players *pre;
struct players *auxi;
auxi = first; //initialization of auxi
while(auxi != NULL) //this should run the loop till the end of the list?
{
if(auxi->idteam == key){ //condition to delete
if(auxi == first) first = auxi->next; //erase for the case the node is the first one
else pre->next = post; //erase in the case the node is anywhere else
}
pre = auxi; //saves the current value of auxi
auxi = auxi->next; //increments the position of auxi
post = auxi->next; //saves the position of the next element
}
}
auxi = auxi->next; //increments the position of auxi
post = auxi->next; //saves the position of the next element
当auxi
变成NULL
时,你最终会做post = (NULL)->next;
,这是一个访问冲突(崩溃)。
你真的不需要post
,只要做:
if(auxi->idteam == key){
if(auxi == first) first = auxi->next;
else pre->next = auxi->next; // We know auxi is not NULL, so this is safe.
}
函数错误
在这个代码片段中
pre = auxi; //saves the current value of auxi
auxi = auxi->next; //increments the position of auxi
post = auxi->next; //saves the position of the next element
声明后 auxi = auxi->next; //increments the position of auxi
auxi可以等于NULL,所以下一条语句
post = auxi->next; //saves the position of the next element
导致未定义的行为。
但这不是唯一的错误。此外,您必须设置正确的节点last
。
你必须释放被删除的节点。
函数可以如下所示
void delete( int key )
{
struct players *prev = NULL;
struct players *auxi = first;;
while ( auxi != NULL )
{
if ( auxi->idteam == key )
{
struct players *tmp = auxi;
if ( auxi == first ) first = auxi->next;
else prev->next = auxi->next;
if ( auxi == last ) last = prev;
auxi = auxi->next;
free( tmp );
}
}