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[英]Why i am not able to assign a struct node pointer with allocated memory to a struct node pointer with previous value of null for making a linked list?
[英]In my Singly linked List implementation why is it even though I allocated memory for the node to be freed, the pointer to the Node isn't NULL?
使用delete_SLL函數,我要刪除此單鏈接列表的標題(head = 4)。 盡管我得到了正確的輸出,但是保存head值的var struct Node *“ temp”不是NULL。 自由函數不喜歡的變量“ temp”是什么? 將節點temp設置為等於列表頭時是否未對節點temp進行Malloc?
碼:
#include <stdio.h>
#include <stdlib.h>
struct Node{
int item;
struct Node* next;
};
struct List{
struct Node* head;
struct Node* tail;
};
int SLL_empty(struct List* lst){
return lst->head == NULL ;
}
//newLst work
struct List newLst(){
struct List lst;
lst.head = NULL;
lst.tail = NULL;
return lst;
}
//Inserts a node to the front of the list[WORKS]
void insert_SLL(struct List* lst, int x){
struct Node* nde = (struct Node*)malloc(sizeof(struct Node));
nde->next = lst->head;
nde->item = x;
if (SLL_empty(lst))
lst->tail=nde;
lst->head = nde;
}
//Deletes a given Node
void delete_SLL(struct List* lst, int x){
struct Node* temp = (struct Node*)malloc(sizeof(struct Node));;
temp = lst->head;
struct Node* prev = NULL;`enter code here`
//If the head has the key
if (temp != NULL && temp->item == x){
lst->head = temp->next;
temp->next = NULL;
free(temp);
}
// stops once the key is found
while(temp != NULL && temp->item != x){
prev = temp;
temp= temp->next;
}
//If not in list
if (temp == NULL) return;
//If middle
if (temp != NULL && temp->item == x){
prev->next = temp->next;
temp->next = NULL;
}
//if at the end
if (temp != NULL && temp->item == lst->tail->item){
lst->tail= prev;
prev->next = NULL;
}
free(temp);
}
int SLL_pop(struct List *list){
struct Node* nde = list->head;
int item = nde->item;
list->head = nde->next;
free(nde);
if (SLL_empty(list))
list->tail = NULL;
return item;
}
int main(int argc, const char * argv[]) {
int i;
struct List list = newLst();
for (i = 0; i < 5; ++i)
insert_SLL(&list, i);
// printf("The length of the linkedLst is: %d\n",SLL_length(&list));
delete_SLL(&list, 4);
while ( list.head != NULL )
printf("Node: %d\n", SLL_pop(&list));
return 0;
}
free()的主要目的是要求操作系統將分配的內存帶回系統。 您可能無法“看到”該消息,但是如果以后嘗試在“臨時”訪問任何元素,則應該會收到錯誤消息。
而程序中的“臨時”只是一個變量。 由於傳遞值的意義,C不需要,也不能將給定的指針更改為NULL。 記住該指針不再有效是程序員的工作。
或者,您可以在每次釋放指針時將其手動設置為NULL。
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