我想删除所有与密钥具有相同idteam的节点,但它将崩溃。。。我知道它也应该释放内存,但无论如何我认为这应该是可行的:
//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
}
}
最佳答案
auxi = auxi->next; //increments the position of auxi
post = auxi->next; //saves the position of the next element
当
auxi
变成NULL
时,您最终将执行post = (NULL)->next;
,这是一种访问冲突(崩溃)。您不需要
post
,只需: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.
}