[英]SegFault on return statement C
我在这段代码中遇到的问题是,当我尝试在第一次迭代中返回addTrash函数中的struct node * head时。 我怀疑这可能是堆栈损坏,但我不确定,我也不知道我如何确定我给您的代码中的错误。
这是一个双向链接的链表,其值仅作为其持有的数据。
struct node * modifyMainList( struct node *head, int link2Delete){
struct node * curr = head;
struct node * temp;
int i = 0;
//traverse list link2Delete amount of times
while(i != link2Delete){
curr = curr -> next;
}
//head case
if(curr -> previous == NULL){
curr = curr -> next;
head = curr;
return head;
}
//tail case
if(curr -> next == NULL){
temp = curr;
curr = curr -> previous;
curr -> next = NULL;
temp -> previous = NULL;
temp -> next = NULL;
free(temp);
return head;
}
curr -> previous -> next = curr -> next;
curr -> next -> previous = curr -> previous;
curr -> previous = NULL;
curr -> next = NULL;
free(curr);
return head;
}
struct node * addTrash(struct node *mainHead, int link2Delete){
struct node * head = NULL;
struct node * curr = mainHead;
struct node * trashCurr = NULL;;
struct node * temp = NULL;
int i = 0;
printf("im in trash before loop\n\n");
for(i = 0; i < link2Delete; i++){
curr = curr -> next;
}
printf("im in trash before head size check\n\n");
if(head == NULL){
printf("im in trash with head == null\n\n");
//head of main list
if(link2Delete == 0){
printf("im in trash link2delete == null\n\n");
curr = curr -> previous;
head = curr;
curr = curr -> next;
curr -> previous = NULL;
curr -> next = NULL;
return head;
}
printf("im in trash before tail case\n\n");
//tail of main list
if(curr -> next == NULL){
printf("im in trash with tail case\n\n");
head = curr;
head -> previous = NULL;
return head;
}
printf("im in trash before after tail case\n\n");
//every other case
//printf("this is the head value: %d\n\n", head -> value);
head = curr;
//printf("im in trash after head = curr\n\n");
head -> previous = NULL;
//printf("im in trash after head -> previous\n\n");
head -> next = NULL;
printf("im in trash after head -> next\n\n");
printf("this is the head value: %d\n\n", head -> value);
return head;
}else{
printf("im in trash inside else\n\n");
trashCurr = head;
while(trashCurr -> next != NULL){
trashCurr = trashCurr -> next;
}
if(link2Delete == 0){
temp = curr;
trashCurr -> next = temp;
temp -> previous = trashCurr;
trashCurr = temp;
trashCurr -> next = NULL;
return head;
}
//tail of main list
if(curr -> next == NULL){
temp = curr;
trashCurr = temp;
temp -> previous = trashCurr;
temp -> next = NULL;
trashCurr -> next = temp;
return head;
}
//every other case
temp = curr;
temp -> previous = trashCurr;
trashCurr -> next = temp;
trashCurr = temp;
trashCurr -> next = NULL;
return head;
}
}
void generateRandom(struct node *mainHead, int size){
int i = 0;
int link2Delete = 0;
struct node *head = NULL;
srand ( time(NULL) );
int number2Delete = rand() % size + 1;
printf("this is the rand number: %d\n", rand());
printf("this is the number of nodes to be deleted: %d\n", number2Delete);
for (i = 0; i < number2Delete; i++) {
// Pick a random node (payload) to delete.
link2Delete = (rand() % size);
printf("this is the number of nodes in the list: %d\n", size);
printf("this is the node to be deleted: %d\n", link2Delete);
size--;
if(link2Delete == 0){
mainHead = modifyMainList(mainHead, link2Delete);
//printf("this is the call return: %d\n\n", addTrash(mainHead, link2Delete) -> value);
head = addTrash (mainHead, link2Delete);
}else{
head = addTrash (mainHead, link2Delete);
mainHead = modifyMainList(mainHead, link2Delete);
}
}
return;
}
在您的代码中
while(i != link2Delete){
curr = curr -> next;
}
如果link2Delete!=0
则是一个无限循环。
而且,不是在释放head case
。
由于循环是无限的,因此curr = curr->next
将继续重复,并将指向某些垃圾指针(如果循环条件为true)。 然后,它具有不确定的行为。 您可能会得到SEGFAULT。
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