Linked List in C: What happens when we assign the current node to a temporary node pointer? -
i have been given task write function reverse doubly-linked list. wrote function , named reverse(). function given below:
struct node* reverse(struct node** head_ref){ struct node *current=*head_ref; struct node *tempnode=(struct node *)malloc(sizeof(struct node)); if(current->next == null){ return current; } while(current->next != null){ //tempnode=current; tempnode->key=current->key; tempnode->data=current->data; tempnode->next=current->next; tempnode->prev=current->prev; current->next=current->prev; current->prev=tempnode->next; current=tempnode->next; } current->next=current->prev; current->prev=null; *head_ref =current; return current; } where struct node** head_ref parameter takes reference of head node &head argument. function works fine. question here instead of doing
tempnode->key=current->key; tempnode->data=current->data; tempnode->next=current->next; tempnode->prev=current->prev; why can't tempnode=current;? when try doing instead function doesn't produce expected result.
tempnode = current assign pointer, means both tempnode , currect pointing same address in memory.
tempnode = curr; // tempnode = 1036; 1020 1028 1036 1044 +------+------+------+------+ | | | curr | | +------+------+------+------+ ^ | tempnode --------- you need tempnode have same values have deep copy, doing.
tempnode->key=current->key; tempnode->data=current->data; tempnode->next=current->next; // copy content of node tempnode->prev=current->prev; // in part of memory 120 .... 1028 1036 1044 +------+------+------+------+------+ | temp | | curr | | | +------+------+------+------+------+ they have same values pointing address.
i can see overwiting pointers, shouldnt becouse loosing address of allocated memory -> cant free anymore.
store temp variable , after remove node, can free memory.
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