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I'm trying to creating linear linked list recursively with c language,
but keep sticking from here and the code is not working with the error "Linker Tools Error LNK2019". Sadly i can't understand what's the matter. Here is my code.
Thanks for your big help in advance.
#include <stdio.h>
#include <stdlib.h>
struct node
{
char num; //Data of the node
struct node *nextptr; //Address of the next node
};
typedef struct node element;
typedef element *link;
link head;
void displayList(); // function to display the list
int main()
{
char s[] = "abc";
link stol(s);
{
link head;
if (s[0] == '\0')return(NULL);
else {
head = (link)malloc(sizeof(element));
head->num = s[0];
head->nextptr = stol(s + 1);
return(head);
}
}
printf("\n\n Linked List : To create and display Singly Linked List :\n");
printf("-------------------------------------------------------------\n");
displayList();
return 0;
}
void displayList()
{
link tmp;
if (head == NULL)
{
printf(" List is empty.");
}
else
{
tmp = head;
while (tmp != NULL)
{
printf(" Data = %d\n", tmp->num); // prints the data of current node
tmp = tmp->nextptr; // advances the position of current node
}
}
}
You redefine a link object called head in your main() function. It hides the global head variable.
Removing the definition inside main would fix your problem, but you should consider passing a link* as a parameter to your displayList function in any case.
I've just noticed this statement return(head); in main(). You program exits prematurely as a result as well.
Everytime I look at your app, I find more issues. If I were you, I'd start by creating a function that adds a node to the list. It's much easier to add new nodes to the front of the list, so you should try that first. Try adding to the tail once you get this running. Adding to the tail is very similar, but you have to 'walkthe list first to get to the last element, exactly as you already do indisplayList()` Another way is keeping the address of the last node* you've added to the list. Like I said, it adds a bit of complexity, so get it working with addToHead first.
void addToHead(link* l, node* n)
{
n->nextptr = l->nextptr;
l->nextptr = n;
}
in your main, you can allocate one new node at a time, as you already do with malloc(). Initialize its contents num with an integer, and let addToHead deal with the pointer stuff. Your use of pointers is terrible, but lists are quite easy, and addToList pretty much shows what can and what should be put in pointers - namely other pointers.
You can remove almost everything in main() before the first printf. You'll have to
start loop:
write a prompt so the user knows what to do using printf()
read input from user using scanf("%d", &n), or equivalent.
break from the loop if user enters a negative value.
malloc() a new node
set its data num = n
call addToHead to add the node.
Loop until user enters an empty string, or -1.
That should take about 8 to 10 lines of code. if in doubt, you will easily find documentation on scanf, with google or on http://en.cppreference.com/w/c.
I'm trying to just reverse a singly linked list, but with a bit of a twist. Rather than having the pointer to the next node be the actual next node, it points to the pointer in that next node.
struct _Node
{
union
{
int n;
char c;
} val;
void *ptr; /* points to ptr variable in next node, not beginning */
int var;
};
typedef struct _Node Node;
I know how to reverse a normal singly linked list and I think I have the general idea of how to go about solving this one, but I'm getting a segfault when I'm trying to access head->ptrand I don't know why.
Node *reverse(Node *head)
{
Node * temp;
Node * prev = NULL;
while(head != NULL)
{
temp = head->ptr + 4; /* add 4 to pass union and get beginning of next node */
head->ptr = prev;
prev = head;
head = temp;
}
return prev;
}
Even if I try and access head->ptr without adding 4, I get a segfault.
The driver that I have for this code is only an object file, so I can't see how things are being called or anything of the sort. I'm either missing something blatantly obvious or there is an issue in the driver.
First, I'll show you a major problem in your code:
while (head) // is shorter than while(head != NULL)
{
// Where does the 4 come from?
// And even if: You have to substract it.
// so, definitively a bug:
// temp = head->ptr + 4; /* add 4 to pass union and get beginning of next node */
size_t offset_ptr = (char*)head->ptr - (char*)head;
// the line above should be moved out of the while loop.
temp = head->ptr - offset_ptr;
Anyways, your algorithm probably won't work as written. If you want to reverse stuff, you are gonna have to work backwards (which is non-trivial in single linked lists). There are two options:
count the elements, allocate an array, remember the pointers in that array and then reassign the next pointers.
create a temporary double linked list (actually you only need another single reversely linked list, because both lists together form a double linked list). Then walk again to copy the next pointer from your temporary list to the old list. Remember to free the temporary list prior to returning.
I tried your code and did some tweaking, well in my opinion your code had some logical error. Your pointers were overwritten again and again (jumping from one node to another and back: 1->2 , 2->1) which were leading to suspected memory leaks. Here, a working version of your code...
Node *reverse(Node *head)
{
Node *temp = 0;
//Re-ordering of your assignment statements
while (head) //No need for explicit head != NULL
{
//Here this line ensures that pointers are not overwritten
Node *next = (Node *)head->ptr; //Type casting from void * to Node *
head->ptr = temp;
temp = head;
head = next;
}
return temp;
}
I've sort of ended up writing a translator from Haskell to C, a hobby thing..
Haskell's (:)-function, with type a -> [a] -> [a] is what I want to do in C.
1 : [2,3] is in fact 1 : (2 : (3 : [])) if I'm not mistaken.
Say I want to create an infinite list with increasing numbers in it:
lst i = i : lst (i + 1)
How do I do this in C? I imagine the final product looking something along the lines of:
int* lst(int i) {
return cons(i, lst(i + 1));
}
My thought so far:
C has arrays.
Arrays need to be of a defined length, this clashes with recursive reasoning.
C has pointers.
Arrays decay to pointers when passed as arguments anyway so, might as well use pure pointers.
array[i] is equivalent to *(ptr + i), I'm thinking I can use this to get around the problem of having to define things you cannot know (final length of the list etc).
I'm unsure of the implementation of cons though.. My best guess is:
int* cons(int head, int *tail) {
int *ptr;
*(ptr + 1) = *tail;
*ptr = head;
return ptr;
}
Pointers are hard for me, dereferencing etc etc, I don't know C very well and my brain hurts. I just want to make a pointer which contains both the head and the tail. Order is not important for the moment.
It compiles, but that's as far as it goes. Help would be appreciated, I'm open to suggestions, I'm not even sure I'm on the right track or if it's even possible.
First, this is what your function is doing:
int* cons(int head, int *tail) {
int *ptr; // Declare a pointer on the stack
*(ptr + 1) = *tail; // Set the int located after the one pointed by the (uninitialised) pointer ptr to the value pointed to by tail
*ptr = head; // Set the value that the (still unitiliased) points to to head
return ptr; // Return an uninitialised value
}
Second, what you want is a linked list. You can create a structure just like data List a = [] | (:) a (List a) in C. For example,
typedef struct list {
void *element;
struct list *next;
} list_t;
Now cons would look like this:
list_t *harp_cons(void *element, list_t *rest) {
list_t *list = (list_t*)malloc(sizeof(struct list_t));
list->element = element;
list->next = rest;
return list;
}
This is allocating data on the heap, so you need to free it afterwards. You can provide a function free_list which would look like the following. (Assuming the elements can simply be freed with free() for the sake of simplicity.)
void free_list(list_t *list) {
if(list != NULL) {
if(list->next != NULL) {
free_list(list->next);
}
free(list->element);
free(list);
}
}
I just took that code from some of my open source code.
If you want to look at a full implementation of (sort of) a list API: https://github.com/thoferon/harp/blob/master/libharp/list.c.
I am having trouble grasping the concepts of struct and the linked list data structure together. For example lets say we have have this code: a struct that has a worker's content and a linked list of these structs that contains nodes of each worker and a pointer to the next node(?).
typedef struct Schedule {
char name[10];
char description[10];
int hours;
int workordernum;
} Work;
typedef struct linkedlist {
struct Schedule work;
struct linkedlist *next;
} Node;
The problem is how do you make a method that always adds a node in the beginning of the list, a method that adds it anywhere(middle) in the list using user defined workordernum, and a method that always puts it in the end.
I am not quite understanding -> and * usages properly. I did read online about creating head and tail nodes but I didn't quite get its usage properly, since they had a struct for a list and a struct for a node.
Another thing which I didn't get is, lets say adding a node to the beginning of the list works, how do you go about then changing every single workordernum value for all the nodes that were previously there?
I understand one must keep track of every time a node gets added, removed, or moved, meaning every time these methods are called we must have a variable that keeps track of the number. So if we have a node in the list all ready, its order would be one, then we add another one to the beginning, how would we change the order number 1 to 2 and the one being added to 1?
Or how does node->next->next->next work if we only have one pointer? Then how would we print all of them? Since we cannot use a for loop.
So these are the concepts I cant grasp code wise. I would KINDLY appreciate it if you take your time to explain it rather than just give the code, if possible. Because I would have to apply what I learn to moving around and deleting nodes as well. I want to learn it on my own. If something must be given as a code example then that's fine but please just don't post all the answer codes for me.
-Thank You
*Please forgive any format errors since I am new to this site.
Edit: I do understand a pointer is an address and that -> pertains to "pointing to" a member. I mean I understand all the basics, but my understanding isn't firm enough else I could be doing what I need help with.
Edit2: I will try and make a head node with linked list from what I have learned so far. I will be using the structs above and it will be loose code, not perfect. This is just to make sure I'm on the right track so far.
int main() {
// creates a work struct to hold user content
Work *workstruct = (Work*)malloc((sizeof(struct Schedule));
// creates first node to hold a work struct for linkedlist
Node *initialNode = (Node*)malloc((sizeof(struct linkedlist));
// Method call to add work nodes to list in main
addWork(initialNode, workstruct);
}
void addWork(Node *initialNode, Work *workstruct) {
// Suppose user already initialized workstruct
// create double-pointer to make initialNode as head
Node **head = (Node **)malloc(sizeof(struct linkedlist));
// assigns user's workstruct to the workstruct of initialNode
initialNode->work = *workstruct;
// If im understanding what you have taught me so far,
// this is how you always add a node on the head
initialNode->next = *head;
*head = initialNode;
}
The only issue I seem to have so far is that every time I try and add a new node to the list, it makes the new node the head but loses the previous node that was in the list.
Linked Lists - 101 - Singly-linked Lists
This is a long answer. The reason I have gone into so much detail is that there are a large number of linked-list questions that I hope to answer in on place, with proper context.
When I learned C, I had a hard time with pointers. However, after implementing a linked list, I finally started to grasp the concept of pointers. Master linked lists is a good thing in C, and it will help you be comfortable with pointers. When things seem confusing, grab a pencil and paper and sketch out a diagram of a list and the associated links to nodes. I occasionally do that when I am working with complex list implementations.
A linked list is a way to store data records. Unlike an array where all elements occupy one contiguous memory block of memory, linked list elements occupy random fragments of memory.
There are two basic types of linked list; a singly-linked list, and a doubly-linked list. The difference is that a singly-linked list can only be traversed in one direction; while a doubly-linked list can be traversed in both directions.
A singly-linked list is accessed via its "head" pointer, or a pointer to the head list node. A doubly-linked list may also be accessed by via its "head" pointer, or via its "tail" pointer.
Unlike an array where each element of the array can be directly addressed by its array index, linked list elements are accessed sequentially.
Here is a layout of a singly-linked list:
Node #1 Node #2 Node #3 EndOfList
---------- ---------- -------- ---------
HEADPTR--> NEXTPTR--> NEXTPTR--> NEXTPTR--> NULL
DataPayload DataPayload DataPayload
Each node in the list, with its data payload, is separately allocated. A node structure (in C) might look something like this:
typedef struct NODE_PAYLOAD_S
{
/* Data Payload (defined by coder) */
char name[10];
char desc[10];
int hours;
int workordernum;
} NODE_PAYLOAD_T;
typedef struct LIST_NODE_S
{
/* Next-node pointer */
struct LIST_NODE_S *next; /* pointer to the next node in the list. */
NODE_PAYLOAD_T payload; /* Data Payload (defined by coder) */
} LIST_NODE_T;
To initialize a singly-linked list of the above structure:
LIST_NODE_T *listHead = NULL;
'listHead' is now a pointer to a linked list (with no nodes).
Here is how to add a new node at the head of the this list:
int LIST_InsertHeadNode(
LIST_NODE_T **IO_head,
Q: Why is a "double-pointer" required here (ie: LIST_NODE_T **...)? Why not use a "single-level' pointer (ie: LIST_NODE_T *...)?
A: A "single" pointer to the list head would not be sufficient for this operation. Specifically, this operation designates a new "head node". This means that the this function needs to modify the pointer that points to the head node.
BEFORE:
Node #Y Node #Z EndOfList
---------- ---------- ---------
HEADPTR--> NEXTPTR--> NEXTPTR--> NULL
DataPayload DataPayload
AFTER:
New Node Node #Y Node #Z EndOfList
---------- ---------- -------- ---------
HEADPTR--> NEXTPTR--> NEXTPTR--> NEXTPTR--> NULL
DataPayload DataPayload DataPayload
Note that before, HEADPTR was pointing to 'Node #Y'; then after, HEADPTR is pointing at 'New node'. When this function is called, the address of the listHead pointer is passed in, allowing this function to change where the listHead pointer is pointing. In otherwords, the address of the listHead pointer is being passed into this function, which is represented (internally to this function) as a pointer to the listHead pointer (a pointer to a pointer). That is why it is a "double-pointer".
char *I__name,
char *I__desc,
int I__hours,
int I__workordernum
)
{
int rCode=0;
LIST_NODE_T *newNode = NULL;
/* Allocate memory for new node (with its payload). */
newNode=malloc(sizeof(*newNode));
if(NULL == newNode)
{
rCode=ENOMEM; /* ENOMEM is defined in errno.h */
fprintf(stderr, "malloc() failed.\n");
goto CLEANUP;
Q: What's this 'goto CLEANUP;' business?
A: The C language, unlike C++ and JAVA, has no concept of an 'exception'. Error checking in C is critical. There are a number of reasons that the malloc() function might fail, and if it does, the code has to handle it as gracefully as possible. The 'goto CLEANUP' statement causes the normal program flow to skip code jumping to the 'CLEANUP:' label (within this function, below).
Obviously, if malloc() has failed, it would be unwise to try to initialize the NULL pointer (returned by the failed malloc) with the lines that immediately follow. So, it is important to divert the flow of the program to skip this initialization (and the linkage that comes later).
There is nothing special about the 'CLEANUP:' label. I could have called it 'ERROR:', 'EXIT:', 'FINISH:', 'LIAHONA:', 'MY_BAD' or anything else that suited my pleasure. (Labels don't have to be uppercase, nor do they have to be placed at the left margin. However, my style is to do that so that they stand out.)
Labels, such as 'CLEANUP:' has a scope that is limited to the boundaries of the function where they are placed; which allows each function to have a unique 'CLEANUP:' label (if needed).
}
/* Initialize the new node's payload. */
snprintf(newNode->payload.name, sizeof(newNode->payload.name), "%s", I__name);
snprintf(newNode->payload.desc, sizeof(newNode->payload.desc), "%s", I__desc);
newNode->payload.hours = I__hours;
newNode->payload.workordernum = I__workordernum;
/* Link this node into the list as the new head node. */
newNode->next = *IO_head;
*IO_head = newNode;
CLEANUP:
return(rCode);
}
The above function might be called as follows:
#include <stdio.h>
#include <errno.h>
int LIST_InsertHeadNode(LIST_NODE_T **, char *, char *, int, int);
int main(void)
{
int rCode=0;
LIST_NODE_T *listHead = NULL;
rCode=LIST_InsertHeadNode(&listHead, "Mahonri", "Jareds Bro", 4, 2421);
if(rCode)
{
fprintf(stderr, "LIST_InsertHeadNode() reports: %d\n", rCode);
goto CLEANUP;
}
CLEANUP:
return(rCode);
}
The LIST_InsertHeadNode() function may be called multiple times. Each call will add a new node to the list. The new node will be placed at the "head" of the list, which has the effect of pushing the rest of the nodes further down the list.
After adding several nodes to the list, it might be good to access the list; perhaps to print the payload of each node:
int PrintListPayloads(
LIST_NODE_T *head;
)
{
int rCode=0;
LIST_NODE_T *cur = head
int nodeCnt=0;
while(cur)
{
++nodeCnt;
printf("%s, %s, %d, %d\n",
cur->payload.name,
cur->payload.desc,
cur->payload.hours,
cur->payload.workordernum
);
cur=cur->next;
}
printf("%d nodes printed.\n", nodeCnt);
return(rCode);
}
The above function can be called from main():
#include <stdio.h>
#include <errno.h>
int LIST_InsertHeadNode(LIST_NODE_T **, char *, char *, int, int);
int PrintListPayloads(LIST_NODE_T *);
int main(void)
{
int rCode=0;
LIST_NODE_T *listHead = NULL;
/* Insert a linked-list node. */
rCode=LIST_InsertHeadNode(&listHead, "Mahonri", "Jareds Bro", 4, 2421);
if(rCode)
{
fprintf(stderr, "LIST_InsertHeadNode() reports: %d\n", rCode);
goto CLEANUP;
}
/* Insert a linked-list node. */
rCode=LIST_InsertHeadNode(&listHead, "Joe", "CEO", 5, 2419);
if(rCode)
{
fprintf(stderr, "LIST_InsertHeadNode() reports: %d\n", rCode);
goto CLEANUP;
}
/* Insert a linked-list node. */
rCode=LIST_InsertHeadNode(&listHead, "Eve", "Mother", 24, 2);
if(rCode)
{
fprintf(stderr, "LIST_InsertHeadNode() reports: %d\n", rCode);
goto CLEANUP;
}
rCode=PrintListPayloads(listHerad);
if(rCode)
{
fprintf(stderr, "PrintListPayloads() reports: %d\n", rCode);
goto CLEANUP;
}
CLEANUP:
return(rCode);
}
Adding nodes at the head of a list [ie: LIST_InsertHeadNode()] is one way to add nodes. However, there are times where adding nodes to the other end of the list (ie: the list 'tail') is preferable. The code below shows how this is done.
First, a function that will return the current 'tail node' of a list.
int LIST_GetTailNode(
LIST_NODE_T *I__listHead, /* The caller supplied list head pointer. */
LIST_NODE_T **_O_listTail /* The function sets the callers pointer to the
last node. */
)
{
int rCode=0;
LIST_NODE_T *curNode = I__listHead;
/* Iterate through all list nodes until the last node is found. */
/* The last node's 'next' field, which is always NULL. */
if(curNode)
{
while(curNode->next)
curNode=curNode->next;
}
/* Set the caller's pointer to point to the last (ie: tail) node. */
if(_O_listTail)
*_O_listTail = curNode;
return(rCode);
}
Next, a function that will insert a node at the tail of the list.
int LIST_InsertTailNode(
LIST_NODE_T **IO_head,
char *I__name,
char *I__desc,
int I__hours,
int I__workordernum
)
{
int rCode=0;
LIST_NODE_T *tailNode;
LIST_NODE_T *newNode = NULL;
/* Get a pointer to the last node in the list. */
rCode=LIST_GetTailNode(*IO_head, &tailNode);
if(rCode)
{
fprintf(stderr, "LIST_GetTailNode() reports: %d\n", rCode);
goto CLEANUP;
}
Important note: The LIST_GetTailNode() function will set the tailNode pointer to the last node in the linked list; -unless- there are no nodes in the list. When the list is empty, LIST_GetTailNode() will set the tailNode pointer to NULL.
/* Allocate memory for new node (with its payload). */
newNode=malloc(sizeof(*newNode));
if(NULL == newNode)
{
rCode=ENOMEM; /* ENOMEM is defined in errno.h */
fprintf(stderr, "malloc() failed.\n");
goto CLEANUP;
}
/* Initialize the new node's payload. */
snprintf(newNode->payload.name, sizeof(newNode->payload.name), "%s", I__name);
snprintf(newNode->payload.desc, sizeof(newNode->payload.desc), "%s", I__desc);
newNode->payload.hours = I__hours;
newNode->payload.workordernum = I__workordernum;
/* Link this node into the list as the new tail node. */
newNode->next = NULL;
if(tailNode)
tailNode->next = newNode;
else
This 'else' case indicates occurs when tailNode is NULL, meaning that (currently) the linked list has no nodes. In this case, this node will be the first (head) node in the list (as well as the last). So, the caller's 'List Head' pointer is updated to indicate that this new node is now the head node.
*IO_head = newNode;
CLEANUP:
return(rCode);
}
The LIST_InsertTailNode() function is called the same way LIST_InsertHeadNode() is called. The only difference is that with LIST_InsertTailNode(), the new node is inserted at the list's tail, rather than at the list's head.
OK, so now you can insert a new node at the head, or tail of the list. What about inserting a new node somewhere in the middle of the list?
For example, assume that you want to have a list where all the nodes are sorted by some field in the payload like 'name'. While it is possible to add all the nodes, and then sort the list afterwords; it is much easier to insert each new node into the list in it's proper place. Doing that, the list will always be maintained in sorted order automatically.
Accomplishing this is done in two steps. First, allocate and initialize the new node. Then figure out where its proper place is in the list, then link the new node into the list at that location.
First, a function that will return what will be the 'parent node' to the new node. (This node assumes that the list is being maintained in sorted order by name):
int LIST_FetchParentNodeByName(
LIST_NODE_T *I__head,
const char *I__name,
LIST_NODE_T **_O_parent
)
{
int rCode=0;
LIST_NODE_T *parent = NULL;
LIST_NODE_T *curNode = I__head;
/* Inform the caller of an 'empty list' condition. */
if(NULL == I__head)
{
rCode=ENOENT;
goto CLEANUP;
}
/* Find a node with a payload->name string greater than the I__name string */
while(curNode)
{
if(strcmp(curNode->payload.name, I__name) > 0)
break;
parent = curNode; /* Remember this node. It is the parent of the next node. */
curNode=curNode->next; /* On to the next node. */
}
/* Set the caller's 'parent' pointer. */
if(_O_parent)
*_O_parent = parent;
CLEANUP:
return(rCode);
}
And now, a function that will insert new nodes, keeping the list sorted by name.
int LIST_InsertNodeByName(
LIST_NODE_T **IO_head,
char *I__name,
char *I__desc,
int I__hours,
int I__workordernum
)
{
int rCode=0;
LIST_NODE_T *parent;
LIST_NODE_T *newNode = NULL;
/* Allocate memory for new node (with its payload). */
newNode=malloc(sizeof(*newNode));
if(NULL == newNode)
{
rCode=ENOMEM; /* ENOMEM is defined in errno.h */
fprintf(stderr, "malloc() failed.\n");
goto CLEANUP;
}
/* Initialize the new node's payload. */
snprintf(newNode->payload.name, sizeof(newNode->payload.name), "%s", I__name);
snprintf(newNode->payload.desc, sizeof(newNode->payload.desc), "%s", I__desc);
newNode->payload.hours = I__hours;
newNode->payload.workordernum = I__workordernum;
/* Find the proper place to link this node */
rCode=LIST_FetchParentNodeByName(*IO_head, I__name, &parent);
switch(rCode)
{
case 0:
break;
case ENOENT:
/* Handle empty list condition */
newNode->next = NULL;
*IO_head = newNode;
rCode=0;
goto CLEANUP;
default:
fprintf(stderr, "LIST_FetchParentNodeByName() reports: %d\n", rCode);
goto CLEANUP;
}
Important note: The LIST_FetchParentNodeByName() function will set the parent pointer to the node in the list that is immediately less than the specified I__name; -unless- the head node is greater than the specified I__name. For this special case, LIST_FetchParentNodeByName() will set the parent pointer to NULL.
/* Handle the case where all current list nodes are greater than the new node. */
/* (Where the new node will become the new list head.) */
if(NULL == parent)
{
newNode->next = *IO_head;
*IO_head = newNode;
goto CLEANUP;
}
/* Final case, insert the new node just after the parent node. */
newNode->next = parent->next;
parent->next = newNode;
CLEANUP:
return(rCode);
}
The LIST_InsertNodeByName() function is called the same way LIST_InsertHeadNode() or LIST_InsertTailNode() is called. The only difference is that with LIST_InsertNodeByName(), the new node is inserted into its sorted (by name) location in the list; rather than at the list head or tail.
There will be occasions when a node will have to be deleted from the list. This is done by locating the node to be deleted, unlinking the node from the list, and then deleting the node and it's payload.
First, a function to locate a specific node by the payload name field.
int LIST_FetchNodeByName(
LIST_NODE_T *I__head,
const char *I__name,
LIST_NODE_T **_O_node,
LIST_NODE_T **_O_parent
)
{
int rCode=0;
LIST_NODE_T *parent = NULL;
LIST_NODE_T *curNode = I__head;
/* Search the list for a matching payload name. */
while(curNode)
{
if(0 == strcmp(curNode->payload.name, I__name))
break;
parent = curNode; /* Remember this node; it will be the parent of the next. */
curNode=curNode->next;
}
/* If no match is found, inform the caller. */
if(NULL == curNode)
{
rCode=ENOENT;
goto CLEANUP;
}
/* Return the matching node to the caller. */
if(_O_node)
*_O_node = curNode;
/* Return parent node to the caller. */
if(_O_parent)
*_O_parent = parent;
CLEANUP:
return(rCode);
}
Here is a function that will delete a node from the list that matches a specific payload name.
int LIST_DeleteNodeByName(
LIST_NODE_T **IO_head,
char *I__name
)
{
int rCode=0;
LIST_NODE_T *parent;
LIST_NODE_T *delNode = NULL;
/* Find the node to delete. */
rCode=LIST_FetchNodeByName(*IO_head, I__name, &delNode, &parent);
switch(rCode)
{
case 0:
break;
case ENOENT:
fprintf(stderr, "Matching node not found.\n");
goto CLEANUP;
default:
fprintf(stderr, "LIST_FetchNodeByName() reports: %d\n", rCode);
goto CLEANUP;
}
Important note: The LIST_FetchNodeByName() function will set the parent pointer of the delNode; -unless- the the delNode is the head node. For this special case, LIST_FetchNodeByName() will set the parent pointer to NULL.
/* Unlink the delNode from the list. */
if(NULL == parent)
*IO_head = delNode->next;
else
parent->next = delNode->next;
/* Free the delNode and its payload. */
free(delNode);
CLEANUP:
return(rCode);
}
NOTE: All code above has been tested and should be functional, and can be downloaded as: 23279119_List_101.c
(To be continued- as per request...)
Is it possible to delete a middle node in the single linked list when the only information available we have is the pointer to the node to be deleted and not the pointer to the previous node?After deletion the previous node should point to the node next to deleted node.
It's definitely more a quiz rather than a real problem. However, if we are allowed to make some assumption, it can be solved in O(1) time. To do it, the strictures the list points to must be copyable. The algorithm is as the following:
We have a list looking like: ... -> Node(i-1) -> Node(i) -> Node(i+1) -> ... and we need to delete Node(i).
Copy data (not pointer, the data itself) from Node(i+1) to Node(i), the list will look like: ... -> Node(i-1) -> Node(i+1) -> Node(i+1) -> ...
Copy the NEXT of second Node(i+1) into a temporary variable.
Now Delete the second Node(i+1), it doesn't require pointer to the previous node.
Pseudocode:
void delete_node(Node* pNode)
{
pNode->Data = pNode->Next->Data; // Assume that SData::operator=(SData&) exists.
Node* pTemp = pNode->Next->Next;
delete(pNode->Next);
pNode->Next = pTemp;
}
Mike.
Let's assume a list with the structure
A -> B -> C -> D
If you only had a pointer to B and wanted to delete it, you could do something like
tempList = B->next;
*B = *tempList;
free(tempList);
The list would then look like
A -> B -> D
but B would hold the old contents of C, effectively deleting what was in B. This won't work if some other piece of code is holding a pointer to C. It also won't work if you were trying to delete node D. If you want to do this kind of operation, you'll need to build the list with a dummy tail node that's not really used so you guarantee that no useful node will have a NULL next pointer. This also works better for lists where the amount of data stored in a node is small. A structure like
struct List { struct List *next; MyData *data; };
would be OK, but one where it's
struct HeavyList { struct HeavyList *next; char data[8192]; };
would be a bit burdensome.
Not possible.
There are hacks to mimic the deletion.
But none of then will actually delete the node the pointer is pointing to.
The popular solution of deleting the following node and copying its contents to the actual node to be deleted has side-effects if you have external pointers pointing to nodes in the list, in which case an external pointer pointing to the following node will become dangling.
I appreciate the ingenuity of this solution (deleting the next node), but it does not answer the problem's question. If this is the actual solution, the correct question should be "delete from the linked list the VALUE contained in a node on which the pointer is given". But of course, the correct question gives you a hint on solution.
The problem as it is formulated, is intended to confuse the person (which in fact has happened to me, especially because the interviewer did not even mention that the node is in the middle).
One approach would be to insert a null for the data. Whenever you traverse the list, you keep track of the previous node. If you find null data, you fix up the list, and go to the next node.
The best approach is still to copy the data of the next node into the node to be deleted, set the next pointer of the node to the next node's next pointer, and delete the next node.
The issues of external pointers pointing to the node to be deleted, while true, would also hold for the next node. Consider the following linked lists:
A->B->C->D->E->F and G->H->I->D->E->F
In case you have to delete node C (in the first linked list), by the approach mentioned, you will delete node D after copying the contents to node C. This will result in the following lists:
A->B->D->E->F and G->H->I->dangling pointer.
In case you delete the NODE C completely, the resulting lists will be:
A->B->D->E->F and G->H->I->D->E->F.
However, if you are to delete the node D, and you use the earlier approach, the issue of external pointers is still there.
The initial suggestion was to transform:
a -> b -> c
to:
a ->, c
If you keep the information around, say, a map from address of node to address of the next node then you can fix the chain the next time to traverse the list. If need to delete multiple items before the next traversal then you need to keep track of the order of deletes (i.e. a change list).
The standard solution is consider other data structures like a skip list.
Maybe do a soft delete? (i.e., set a "deleted" flag on the node) You can clean up the list later if you need to.
Not if you want to maintain the traversability of the list. You need to update the previous node to link to the next one.
How'd you end up in this situation, anyway? What are you trying to do that makes you ask this question?
You'll have to march down the list to find the previous node. That will make deleting in general O(n**2). If you are the only code doing deletes, you may do better in practice by caching the previous node, and starting your search there, but whether this helps depends on the pattern of deletes.
Given
A -> B -> C -> D
and a pointer to, say, item B, you would delete it by
1. free any memory belonging to members of B
2. copy the contents of C into B (this includes its "next" pointer)
3. delete the entire item C
Of course, you'll have to be careful about edge cases, such as working on lists of one item.
Now where there was B, you have C and the storage that used to be C is freed.
Considering below linked list
1 -> 2 -> 3 -> NULL
Pointer to node 2 is given say "ptr".
We can have pseudo-code which looks something like this:
struct node* temp = ptr->next;
ptr->data = temp->data;
ptr->next = temp->next;
free(temp);
yes, but you can't delink it. If you don't care about corrupting memory, go ahead ;-)
Yes, but your list will be broken after you remove it.
In this specific case, traverse the list again and get that pointer! In general, if you are asking this question, there probably exists a bug in what you are doing.
In order to get to the previous list item, you would need to traverse the list from the beginning until you find an entry with a next pointer that points to your current item. Then you'd have a pointer to each of the items that you'd have to modify to remove the current item from the list - simply set previous->next to current->next then delete current.
edit: Kimbo beat me to it by less than a minute!
You could do delayed delinking where you set nodes to be delinked out of the list with a flag and then delete them on the next proper traversal. Nodes set to be delinked would need to be properly handled by the code that crawls the list.
I suppose you could also just traverse the list again from the beginning until you find the thing that points to your item in the list. Hardly optimal, but at least a much better idea than delayed delinking.
In general, you should know the pointer to the item you just came from and you should be passing that around.
(Edit: Ick, with the time it took me to type out a fullish answer three gazillion people covered almost all the points I was going to mention. :()
The only sensible way to do this is to traverse the list with a couple of pointers until the leading one finds the node to be deleted, then update the next field using the trailing pointer.
If you want to delete random items from a list efficiently, it needs to be doubly linked. If you want take items from the head of the list and add them at the tail, however, you don't need to doubly link the whole list. Singly link the list but make the next field of the last item on the list point to the first item on the list. Then make the list "head" point to the tail item (not the head). It is then easy to add to the tail of the list or remove from the head.
You have the head of the list, right? You just traverse it.
Let's say that your list is pointed to by "head" and the node to delete it "del".
C style pseudo-code (dots would be -> in C):
prev = head
next = prev.link
while(next != null)
{
if(next == del)
{
prev.link = next.link;
free(del);
del = null;
return 0;
}
prev = next;
next = next.link;
}
return 1;
The following code will create a LL, n then ask the user to give the pointer to the node to be deleted. it will the print the list after deletion
It does the same thing as is done by copying the node after the node to be deleted, over the node to be deleted and then delete the next node of the node to be deleted.
i.e
a-b-c-d
copy c to b and free c so that result is
a-c-d
struct node
{
int data;
struct node *link;
};
void populate(struct node **,int);
void delete(struct node **);
void printlist(struct node **);
void populate(struct node **n,int num)
{
struct node *temp,*t;
if(*n==NULL)
{
t=*n;
t=malloc(sizeof(struct node));
t->data=num;
t->link=NULL;
*n=t;
}
else
{
t=*n;
temp=malloc(sizeof(struct node));
while(t->link!=NULL)
t=t->link;
temp->data=num;
temp->link=NULL;
t->link=temp;
}
}
void printlist(struct node **n)
{
struct node *t;
t=*n;
if(t==NULL)
printf("\nEmpty list");
while(t!=NULL)
{
printf("\n%d",t->data);
printf("\t%u address=",t);
t=t->link;
}
}
void delete(struct node **n)
{
struct node *temp,*t;
temp=*n;
temp->data=temp->link->data;
t=temp->link;
temp->link=temp->link->link;
free(t);
}
int main()
{
struct node *ty,*todelete;
ty=NULL;
populate(&ty,1);
populate(&ty,2);
populate(&ty,13);
populate(&ty,14);
populate(&ty,12);
populate(&ty,19);
printf("\nlist b4 delete\n");
printlist(&ty);
printf("\nEnter node pointer to delete the node====");
scanf("%u",&todelete);
delete(&todelete);
printf("\nlist after delete\n");
printlist(&ty);
return 0;
}
void delself(list *list)
{
/*if we got a pointer to itself how to remove it...*/
int n;
printf("Enter the num:");
scanf("%d",&n);
while(list->next!=NULL)
{
if(list->number==n) /*now pointer in node itself*/
{
list->number=list->next->number; /*copy all(name,rollnum,mark..)
data of next to current, disconnect its next*/
list->next=list->next->next;
}
list=list->next;
}
}
If you have a linked list A -> B -> C -> D and a pointer to node B. If you have to delete this node you can copy the contents of node C into B, node D into C and delete D. But we cannot delete the node as such in case of a singly linked list since if we do so, node A will also be lost. Though we can backtrack to A in case of doubly linked list.
Am I right?
void delself(list *list)
{
/*if we got a pointer to itself how to remove it...*/
int n;
printf("Enter the num:");
scanf("%d",&n);
while(list->next!=NULL)
{
if(list->number==n) /*now pointer in node itself*/
{
list->number=list->next->number;
/*copy all(name,rollnum,mark..) data of next to current, disconect its next*/
list->next=list->next->next;
}
list=list->next;
}
}
This is a piece of code I put together that does what the OP was asking for, and can even delete the last element in the list (not in the most elegant way, but it gets it done). Wrote it while learning how to use linked lists. Hope it helps.
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <string>
using namespace std;
struct node
{
int nodeID;
node *next;
};
void printList(node* p_nodeList, int removeID);
void removeNode(node* p_nodeList, int nodeID);
void removeLastNode(node* p_nodeList, int nodeID ,node* p_lastNode);
node* addNewNode(node* p_nodeList, int id)
{
node* p_node = new node;
p_node->nodeID = id;
p_node->next = p_nodeList;
return p_node;
}
int main()
{
node* p_nodeList = NULL;
int nodeID = 1;
int removeID;
int listLength;
cout << "Pick a list length: ";
cin >> listLength;
for (int i = 0; i < listLength; i++)
{
p_nodeList = addNewNode(p_nodeList, nodeID);
nodeID++;
}
cout << "Pick a node from 1 to " << listLength << " to remove: ";
cin >> removeID;
while (removeID <= 0 || removeID > listLength)
{
if (removeID == 0)
{
return 0;
}
cout << "Please pick a number from 1 to " << listLength << ": ";
cin >> removeID;
}
removeNode(p_nodeList, removeID);
printList(p_nodeList, removeID);
}
void printList(node* p_nodeList, int removeID)
{
node* p_currentNode = p_nodeList;
if (p_currentNode != NULL)
{
p_currentNode = p_currentNode->next;
printList(p_currentNode, removeID);
if (removeID != 1)
{
if (p_nodeList->nodeID != 1)
{
cout << ", ";
}
cout << p_nodeList->nodeID;
}
else
{
if (p_nodeList->nodeID !=2)
{
cout << ", ";
}
cout << p_nodeList->nodeID;
}
}
}
void removeNode(node* p_nodeList, int nodeID)
{
node* p_currentNode = p_nodeList;
if (p_currentNode->nodeID == nodeID)
{
if(p_currentNode->next != NULL)
{
p_currentNode->nodeID = p_currentNode->next->nodeID;
node* p_temp = p_currentNode->next->next;
delete(p_currentNode->next);
p_currentNode->next = p_temp;
}
else
{
delete(p_currentNode);
}
}
else if(p_currentNode->next->next == NULL)
{
removeLastNode(p_currentNode->next, nodeID, p_currentNode);
}
else
{
removeNode(p_currentNode->next, nodeID);
}
}
void removeLastNode(node* p_nodeList, int nodeID ,node* p_lastNode)
{
node* p_currentNode = p_nodeList;
p_lastNode->next = NULL;
delete (p_currentNode);
}
Void deleteMidddle(Node* head)
{
Node* slow_ptr = head;
Node* fast_ptr = head;
Node* tmp = head;
while(slow_ptr->next != NULL && fast_ptr->next != NULL)
{
tmp = slow_ptr;
slow_ptr = slow_ptr->next;
fast_ptr = fast_ptr->next->next;
}
tmp->next = slow_ptr->next;
free(slow_ptr);
enter code here
}