C insert element into ascending linked list - c

I have problems with my insert method, as for some reason I end up with infinite loop. Here is my struct:
struct List {
char element;
struct List *next;
};
And here is my insert method:
void insert(struct List *first, char el){
struct List *new=NULL;
struct List *current = first;
new = (struct List*) malloc (sizeof(struct List));
new->element = el;
new->next = NULL;
if (first == NULL){
first = new;
return;
}
while (1){ //this loop never ends
if (current->next == NULL) break;
if (current->next->element < el){
current = current->next;
}else{
break;
}
}
struct List *ex_next = current->next;
current->next = new;
new->next = ex_next;
}
I am aware of similar question here: C - Inserting into linked list in ascending order but it didn't really help me.

The first argument to insert is a pointer. But you need a pointer to a pointer (struct List **first).
If the list is empty, you pass the VALUE NULL into the function (the variable first inside the method have the value NULL). Then you assign a new malloced value to it and return. The variable on the calling side haven't changed and your memory leaked.
When you pass a pointer of a pointer, the variable first holds the address of the variable of the calling method. This way, you can reassign it's value.
Pointers, Pointers of Pointers, Pointers of Pointers of arrays of functions returning function pointers .... thats the fun part of C ;)

Related

C: From char array to linked list

I'm still learning how to program in C and I've stumbled across a problem.
Using a char array, I need to create a linked list, but I don't know how to do it. I've searched online, but it seems very confusing. The char array is something like this char arr[3][2]={"1A","2B","3C"};
Have a look at this code below. It uses a Node struct and you can see how we iterate through the list, creating nodes, allocating memory, and adding them to the linked list. It is based of this GeeksForGeeks article, with a few modifications. I reccommend you compare the two to help understand what is going on.
#include <stdio.h>
#include <stdlib.h>
struct Node {
char value[2];
struct Node * next;
};
int main() {
char arr[3][2] = {"1A","2B","3C"};
struct Node * linked_list = NULL;
// Iterate over array
// We calculate the size of the array by using sizeof the whole array and dividing it by the sizeof the first element of the array
for (int i = 0; i < sizeof(arr) / sizeof(arr[0]); i++) {
// We create a new node
struct Node * new_node = (struct Node *)malloc(sizeof(struct Node));
// Assign the value, you can't assign arrays so we do each char individually or use strcpy
new_node->value[0] = arr[i][0];
new_node->value[1] = arr[i][1];
// Set next node to NULL
new_node->next = NULL;
if (linked_list == NULL) {
// If the linked_list is empty, this is the first node, add it to the front
linked_list = new_node;
continue;
}
// Find the last node (where next is NULL) and set the next value to the newly created node
struct Node * last = linked_list;
while (last->next != NULL) {
last = last->next;
}
last->next = new_node;
}
// Iterate through our linked list printing each value
struct Node * pointer = linked_list;
while (pointer != NULL) {
printf("%s\n", pointer->value);
pointer = pointer->next;
}
return 0;
}
There are a few things the above code is missing, like checking if each malloc is successful, and freeing the allocated memory afterwards. This is only meant to give you something to build off of!

Why am I getting segmentation fault on my code?

I am trying to add an item to the linked list by traversing the list to create the next node. and the last node in the list to point to the newly created node. But I am getting a core dump segmentation fault on it.
void linked_list_add(list_node_t *head, void *item)
{
list_node_t *temp = head;
while(temp->next != NULL)
{
temp = temp->next;
}
list_node_t *new_node = (list_node_t *)malloc(sizeof(list_node_t));
new_node->data = item;
new_node->next = NULL;
new_node->prev = temp;
//if(temp != NULL)
// temp->next = new_node;
// new_node->prev = temp;
}
TEST_F(LinkedList, Add)
{
int i = 3;
linked_list_add(list, &i);
ASSERT_EQ(list->next->data, &i);
i = 4;
linked_list_add(list, &i);
ASSERT_EQ(list->prev->data, &i);
i = 5;
linked_list_add(list, &i);
ASSERT_EQ(list->next->data, &i);
}
This is an answer to summarize the comments.
There are likely at least 3 issues with the code as written:
When the code void linked_list_add(list_node_t *head, void *item) is passed arguments, you generally want to be able to handle a NULL pointer for head. It looks like the while loop immediately goes into searching for the end of the list even if the head is null.
The newly added node, new_node gets the prev pointer updated so that the backwards searchs will be and shouldn't segfault. However, the forward searching isn't preserved. By this I mean that the last non-NULL node in the linked list doesn't have the next pointer pointing to the new_node.
The test ASSERT_EQ(list->prev->data, &i); is likely accessing either a random memory location or a NULL pointer. Given that the OP didn't post the declaration of the list struct it is difficult to say what the default values are/will be. However, unless this list is circular, the value of list->prev is an uninitialized pointer. Depending on your setup (e.g. if there is setup code for the linked list that sets the pointers to null, you could be accessing a NULL pointer there too.
I hope this helps the OP solve their coding problem(s).

Pointer is not being passed during a linked list swap

This will most likely seem like I am missing something obvious but when I try to pass a Linked List pointer to my Selection sort I get a NULL pointer problem. In my C code I have this as my linked list:
typedef struct iorb
{
int base_pri;
struct iorb *link;
char filler[100];
} IORB;
Then I pass my new linked list into this function after creating it:
void swapNodes(POINTER *head, POINTER CurrentHead, POINTER CurrentMinimum, POINTER TempSwap);
POINTER SortList(POINTER *head, char *SortMethod[])
{
POINTER TempHead = *head;
//Only one node, no need to sort.
if (TempHead->link == NULL)
{
return head;
}
//Store node with the new minimum value.
POINTER TempMin = *head;
//Store curent node for swapping
POINTER TempSwap = *head;
//Transverse the list.
POINTER TempCurrent;
for (TempCurrent = *head; TempCurrent->link != NULL; TempCurrent = TempCurrent->link)
{
//Check if this node has a lower priority than the current minimum and if so, swap them.
if (TempCurrent->link->base_pri < TempMin->base_pri)
{
TempMin = TempCurrent->link;
TempSwap = TempCurrent;
}
}
//Swap nodes if the head is not the same as the minimum.
if (TempMin != TempHead)
{
swapNodes(&TempHead, TempHead, TempMin, TempSwap);
}
//Recursively sort the rest of the list.
//FOR SOME REASON THE NODE POINTER IS NOT BEING PASSED HERE (EMPTY)
TempHead->link = SortList(TempHead->link, *SortMethod);
return head;
}
void swapNodes(POINTER *head, POINTER CurrentHead, POINTER CurrentMinimum, POINTER TempSwap)
{
//Set new head as the minimum.
*head = CurrentMinimum;
//Link the current temp swap to the head.
TempSwap->link = CurrentHead;
//Swap pointers.
POINTER temp = CurrentMinimum->link;
CurrentMinimum->link = CurrentHead->link;
CurrentHead->link = temp;
}
I am not sure why it is isn't being passed back into the same function, when I debug the linked list seems okay. I suspect that I am missing something in the swap node function, but I don't quite understand what this is. Can someone please offer some insight as to how this code should go for swapping the nodes around?
If you need additional information please let me know.
SortList(TempHead->link, *SortMethod); needs to be listed as SortList(&TempHead, *SortMethod);
To correctly pass the pointer.

C-modifying the address of a global variable inside a void function

Working on a linked list where I'm storing a pointer to head as a global variable. My question is: How can I assign a (new)value (in this case address) to head from within a void function?
EDIT
Alright here's some code:
Note: this is not how i would design this, but we're not supposed to stray from the spec.
typedef struct NODE Node;
typedef Node *NodePtr;
struct NODE{
char *item;
Node *next;
};
NodePtr first = NULL; //global
//insert function
Boolean insert( char *new_string ) {
printf("insert called\n");
if(first == NULL) {
first = malloc(sizeof(Node));
first->next = NULL;
first->item = new_string;
}
else {
NodePtr inserted = malloc(sizeof(Node));
inserted->next = first;
inserted->item = new_string;
first = inserted;
}
return 1;
}
the problem is that when i use insert() in a function. it works fine. i get a working list. but when another function tries to access first, its empty.
Am I right in assuming that modifying the global variables within the function does not alter the actual first ptr? I know I could pass in a pointer to first as an argument, but I'm not supposed to modify the prototype design.
You'll need to pass in the pointer and a pointer to pointer to head.. Update pointer to head, and head-> next to old head
if it is a global variable just assign the value like you would have done if it was a local variable in your void function.
assuming your struct is similar to this:
struct list {
int data;
struct list* next;
};
and you have the global variable
struct list* L
you can make a new node by making a new node, making the next variable to the head, and making your global value point to the new node:
newnode->next = L;
L = newnode

Doubly linked lists in C

#include<stdio.h>
typedef struct Node
{
int data;
struct Node *next;
struct Node *prev;
} node;
void insert(node *pointer, int data)
{
while(pointer->next!=NULL)
{
pointer = pointer -> next;
}
pointer->next = (node *)malloc(sizeof(node));
(pointer->next)->prev = pointer;
pointer = pointer->next;
pointer->data = data;
pointer->next = NULL;
}
int main()
{
node *start;
start = (node *)malloc(sizeof(node));
int data;
scanf("%d",&data);
insert(start,data);
}
Well, I'm trying to understand the basics of lists in C. I have one question here - the 3rd line from the insert()'s bottom - what is this for? It seems like if the first list's element remains empty and the data is being saved into the second one. But only this works.
In main() there's being created first empty element, right?
while() is not executed as the element is null.
Then the second element is being created. (pointer->null)
Pointer pointing to the first element is set to point to the second element (that 3rd line from the bottom)
And the data is being saved to that second element.
Where do I make a mistake?
pointer = pointer->next;
This line changes the current node we're concentrating on from the last node of the original list to the newly allocated last node of the new list (i.e. the node after the last node of the original list). We then set that node's values directly in the next two lines.
You could get rid of this line and change the two lines below it to read
pointer->next->data = data;
pointer->next->next = NULL;
and you would have the same result.
Edit: Looking further into things, I see more issues:
1) You need #include <stdlib.h> to use malloc().
2) You need to explicitly set start->next = NULL; before you call insert().
In your insert function, you have this assignment:
pointer = pointer->next;
This will not work, as the pointer pointer is passed by value. This means that all changes to the value will be lost when the function returns.
You can pass the pointer by reference:
void insert(node **pointer, int data)
{
/* ... */
*pointer = (*pointer)->next;
/* ... */
}
Or return the pointer from the function:
node *insert(node *pointer, int data)
{
/* ... */
return pointer;
}
There is also the problem that you don't initialize the pointers in the node structure. This means that when you allocate a node structure, the fields in it will be pointing to seemingly random locations.
This is solved simply by setting the next and prev pointer to NULL directly after allocation:
start = malloc(sizeof(node));
start->next = NULL;
start->prev = NULL;

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