I'm new to C and I'm stuck with the insert function in a linked list. When I try printing the list. The result isn't what I expect. I know it has something to do with pointers but I just can't get my head around it. What am I doing wrong here?
#include <stdio.h>
#include <stdlib.h>
#include <assert.h>
#include <string.h>
typedef struct CELL_NODE CellNode;
struct CELL_NODE {
int row;
int column;
CellNode *next;
};
struct LinkedList {
CellNode *head;
};
typedef struct LinkedList LinkedList;
void printList(LinkedList *myList) {
CellNode *curr = (*myList).head;
if (curr != NULL) {
printf("(%d,%d)", (*curr).row, (*curr).column);
if ((*curr).next != NULL) {
curr = (*curr).next;
printf(" - (%d,%d)", (*curr).row, (*curr).column);
}
} else {
printf("The list is empty");
}
printf("\n");
}
void insert(LinkedList *myList, CellNode *node) {
CellNode *ref = (*myList).head;
if (ref == NULL) {
(*myList).head = node;
} else {
while ((*ref).next != NULL) {
ref = (*ref).next;
}
(*ref).next = node;
}
}
int main(int argc, char *argv[]) {
LinkedList myList = { NULL };
for (int k = 0; k < 2; k++) {
CellNode myNode = { 1, k, NULL };
insert(&myList, &myNode);
printList(&myList);
printf("\n");
}
return 1;
}
The result I get is:
(1,0)
(1,1) - (1,1)
I'm expecting:
(1,0)
(1,0) - (1,1)
You should first change every instance of (*x).y to x->y to make your code much more readable.
Then, look at this code:
int main(int argc, char *argv[])
{
LinkedList myList = {NULL};
for(int k = 0 ; k<2 ; k++) {
CellNode myNode = {1,k,NULL};
insert(&myList,&myNode);
printList(&myList);
printf("\n");
}
return 1;
}
You create myNode as a local variable inside the for loop. That means that each iteration of the loop gets a new instance of myNode, destroying the previous one. So you've connected myNode to your linked list through pointers, and then you let it get destroyed the next time through the for loop.
If you're going to let some piece of code stash a pointer to something, you must ensure that something remains valid until there is no longer any possibility of those pointers being dereferenced.
You need to make a decision -- what will own the objects that the linked list contains pointers to? When will that lifetime end? And when they end, what will destroy them?
You haven't done this. So you have objects whose lifetimes end too early.
With
CellNode myNode = {1,k,NULL};
insert(&myList,&myNode);
you are passing a pointer to a local variable. The life time of this variable is just as long as the respective iteration of the loop, i.e. in the second iteration, the object of the first iteration is out of scope. So you will access an object which's life time has already ended by the pointer you stored in your list. This yields undefined behaviour.
Use dynamically generated objects instead (and don't forget to free them later on):
CellNode *myNode = malloc(sizeof(CellNode));
myNode->row = ...
You repeatedly insert a node into the linked list from a local variable that immediately goes out of scope. The behavior is undefined, your program might fail in many unpredictable ways.
You should modify the code this way:
change the insert function to take the element data as arguments and allocate a new node with malloc().
make printList() print the full list, not just the first couple of cells.
change the clumsy (*pointer).member notation into the equivalent but more idiomatic pointer->member notation.
return 0 from main for successful operation.
Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
typedef struct CellNode CellNode;
struct CellNode {
int row;
int column;
CellNode *next;
};
typedef struct LinkedList LinkedList;
struct LinkedList {
CellNode *head;
};
void printList(LinkedList *myList) {
CellNode *curr = myList->head;
if (curr != NULL) {
printf("(%d,%d)", curr->row, curr->column);
while (curr->next != NULL) {
curr = curr->next;
printf(" - (%d,%d)", curr->row, curr->column);
}
} else {
printf("The list is empty");
}
printf("\n");
}
CellNode *insert(LinkedList *myList, int row, int column) {
CellNode *node = malloc(sizeof(*node));
CellNode *ref = myList->head;
if (node != NULL) {
if (ref == NULL) {
myList->head = node;
} else {
while (ref->next != NULL) {
ref = ref->next;
}
ref->next = node;
}
}
return node; // return node pointer to allow the caller to detect out of memory error
}
int main(int argc, char *argv[]) {
LinkedList myList = { NULL };
CellNode *node;
for (int k = 0; k < 2; k++) {
insert(&myList, 1, k);
printList(&myList);
printf("\n");
}
// free the nodes
while ((node = myList->head) != NULL) {
myList->head = node->next;
free(node);
}
return 0;
}
Related
I am trying basic creation of linked list using C. I have written the following code which is working up until first node but fails eventually on second one. I think the issue is where I am trying to display the node values in list separated by arrow(->). I think my logic is right but please correct me. Thanks in advance
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
struct node
{
int number;
struct node *next;
};
typedef struct node NODE;
NODE *node1, *node2, *start, *save;
int main()
{
node1 = (NODE *)malloc(sizeof(NODE));
int i = 0;
start = NULL;
for(i = 0; i < 3; i++)
{
int inf;
printf("Enter node value:");
scanf("%d", &inf);
node1->number = inf;
node1->next = NULL;
if(start == NULL)
{
start = node1;
save = node1;
}
else
{
// save=start;
// start=node1;
// node1->next=save;
node1->next = start;
start = node1;
}
while(node1 != NULL)
{
printf("%d ->",node1->number);
node1 = node1->next;
}
}
return 0;
}
The issues are
How you're allocating your nodes for insertion (i.e. save for one, you're not).
How they're placed in the list once you fix the above.
Don't cast malloc in C programs (read here for why).
Fail to check the success of your scanf invoke.
Fail to check the success of your malloc invoke
Before you get discouraged, things you did correctly:
Did not mask a node pointer in a typedef
Properly included a MCVE for review
Prospected the things you may be doing wrong.
A very simple example of iterating three values into a linked list would look something like this:
#include <stdio.h>
#include <stdlib.h>
struct node
{
int number;
struct node *next;
};
typedef struct node NODE;
int main()
{
NODE *head = NULL, *p;
int i = 0;
for(i = 0; i < 3; i++)
{
int inf;
printf("Enter node value:");
if (scanf("%d", &inf) == 1)
{
p = malloc(sizeof *p);
if (p != NULL)
{
p->number = inf;
p->next = head;
head = p;
}
else
{
perror("Failed to allocate new node");
return EXIT_FAILURE;
}
}
else
{
// failed to read data. break
break;
}
// report current linked list
printf("%d", p->number);
for (p=p->next; p; p = p->next)
printf(" -> %d", p->number);
fputc('\n', stdout);
}
// cleanup the linked list
while (head)
{
p = head;
head = head->next;
free(p);
}
head = NULL;
return 0;
}
Input
The values 1 2 3 are input upon being prompted:
Output
Enter node value:1
1
Enter node value:2
2 -> 1
Enter node value:3
3 -> 2 -> 1
Best of luck.
You should use malloc() inside for loop.
Since it is outside, same memory is being used.
As said by Vamsi, you should use malloc to put the nodes on the heap. You also generally shouldn't cast the output of malloc, it isn't needed. And then you could play around with making a doubly-linked list, where you also have a prev pointer inside your struct.
I cannot figure out how to run this correctly, gives segmentation error. A piece of code is below. Can you look at head too , i am not sure if it is right way of initialising head to null in another file , it is run as follows :
Table tb ;
tb= initialise_table (table_size);
tb = insert(text_words,tb);
//these 3 typedef declarations are in a "some.h" file
typedef struct node * tree_ptr;
typedef char* Key_Type;
typedef struct table* Table;
struct node {
Key_Type element;
tree_ptr left;
tree_ptr right;
};
struct table {
tree_ptr head;
};
Table init_table() {
Table head = NULL;
}
Table insert(Key_Type key ,Table temp ) {
tree_ptr t = (tree_ptr)malloc(sizeof(tree_ptr));
t->element = key;
// t->left = t->right = NULL;
if (temp->head==NULL) {
temp = (Table)malloc (sizeof (Table));
temp->head = t;
printf("empty tree ");
}
else {
temp = insert(t->element,temp);
printf("inserted into ");
}
return temp;
printf("wowo!");
}
The primary issue is in the code which, you say, is used to invoke the functions:
Table tb;
tb = insert(text_words, tb);
You have an uninitialized pointer, tb, which you pass to the function. Inside the function, you have:
Table insert(Key_Type key, Table temp)
{
tree_ptr t = (tree_ptr)malloc(sizeof(*t)); // Fixed size
t->element = key;
// t->left = t->right = NULL;
if (temp->head==NULL)
{
You're therefore accessing (dereferencing) the undefined pointer, and your program is crashing.
You should, I assume, be initializing your table with table_init(), but that function is actually no help whatsoever. It defines and initializes a local variable, but doesn't return anything even though it promises to do so.
Please see Is it a good idea to typedef pointers? The short answer is 'No, it usually isn't a good idea'.
You still have problems even if you fix the calling code like this (a necessary but not sufficient step):
Table tb = NULL;
tb = insert(text_words, tb);
or maybe:
Table tb = init_table();
tb = insert(text_words, tb);
but you need a seriously upgraded version of init_table(), such as:
Table init_table(void)
{
Table root = malloc(sizeof(*head));
root->head = NULL;
return root;
}
Your code in insert() needs to ensure that it does not dereference a null pointer (instead of an indeterminate pointer).
Table insert(Key_Type key, Table root)
{
tree_ptr t = (tree_ptr)malloc(sizeof(*t)); // Fixed size
t->element = key;
t->left = t->right = NULL;
if (root == NULL)
{
root = init_table();
root->head = t;
}
else
{
…
}
return root;
}
Given the Key_Type is a char * in disguise, you may need to review how you save the keys in the tree structure; you may need to use strdup() to copy the data. It is impossible to say for sure without seeing how you are managing the strings that you pass to the insert() function. It could be OK to just save the pointer if the calling code ensures that a new pointer is passed each time. OTOH, if the same pointer is passed each time, you definitely need to copy the data, and using strdup() is a sensible way of doing that. Note that strdup() is standard on POSIX; it is not part of standard C.
Here's one major problem:
tree_ptr t = (tree_ptr) malloc(sizeof(tree_ptr));
should be:
tree_ptr t = (tree_ptr) malloc(sizeof(struct node));
Your code doesn't actually do any binary search. Indeed, it just infinitely recurses creating new nodes. Try something more like this:
#include <stddef.h>
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
typedef struct Node
{
char *element;
struct Node *left;
struct Node *right;
} Node;
typedef struct
{
Node *root;
size_t size;
} Tree;
void Tree_init(Tree *t);
Node *Tree_insert(Tree *t, const char *key);
void Tree_insert_r(Node *subtree, Node *n, size_t size);
void Tree_pre_order_r(Node *subtree);
void Tree_init(Tree *t)
{
t->root = NULL;
t->size = 0;
}
Node *Tree_insert(Tree *t, const char *key)
{
Node *ret = (Node*) malloc(sizeof(Node));
if (ret)
{
ret->left = ret->right = NULL;
if ((ret->element = strdup(key))) /* make a copy of key */
{
if (NULL != t->root)
Tree_insert_r(t->root, ret, t->size);
else
t->root = ret;
++t->size;
}
else
{
free(ret);
ret = NULL;
}
}
return ret;
}
void Tree_insert_r(Node *subtree, Node *n, size_t size)
{
int cmp = strcmp(n->element, subtree->element);
if (cmp < 0 || (cmp == 0 && size % 2 == 0))
{
if (NULL != subtree->left)
subtree = subtree->left;
else
{
subtree->left = n;
return;
}
}
else
{
if (NULL != subtree->right)
subtree = subtree->right;
else
{
subtree->right = n;
return;
}
}
Tree_insert_r(subtree, n, size);
}
void Tree_pre_order_r(Node *subtree)
{
if (NULL == subtree)
return;
fprintf(stdout, "'%s'\n", subtree->element);
Tree_pre_order_r(subtree->left);
Tree_pre_order_r(subtree->right);
}
int main()
{
Tree t;
Tree_init(&t);
Tree_insert(&t, "Hello");
Tree_insert(&t, "World!");
Tree_insert(&t, "etc.");
Tree_pre_order(t.root);
return 0;
}
I'm trying to implement sequence_insert_at using the add_to_front function here
Everything before
typedef struct sequence *Sequence;
is pasted from another c file.
void sequence_insert_at(Sequence s, int pos, int item)
{
struct node* temp = s->lst;
for(; pos > 0; --pos)
{
temp = temp->rest;
}
add_to_front(&temp, item);
++s->length;
if(!temp->rest)
{
s->end = temp;
}
//s->lst = temp;
}
I don't know why I keep getting a runtime error. if I clone s->lst and traverse the clone, I'm not modifying the pointer to the node in s, but if I change temp, s->lst should have the reflected changes since the nodes are all linked still. Any ideas as to how to fix this? I tried creating another node that is one before the temp after traversal, and then setting it->rest = temp, but that failed as well.
following mistakes a could spot but only so far to get the main function run
new_sequence does not initialize anything in Sequence it creates. lst is not initialized when you access it in sequence_insert_at
struct node* temp = s->lst;
here how it should look like
Sequence new_sequence()
{
Sequence s = malloc(sizeof(struct sequence));
if(!s)
{
printf("Out of memory. Can't allocate s\n");
exit(EXIT_FAILURE);
}
s->lst = malloc(sizeof(struct node));
if(! s->lst) {
printf("Out of memory. Can't allocate lst\n");
}
s->lst->rest = NULL;
s->length = 0;
return s;
}
also s->lst->rest has to be set to NULL, this is what tells that the list has no more elements an not end witch turns obsolete.
struct sequence
{
struct node* lst;
int length;
};
You should be passing the sequence itself to your functions not a pointer to some internal data in the sequence.
add_to_front(&temp, item);
Your sequence_insert_at function should be the one that can handle any position not add_to_front() so it is easier to call with the position 0 from add_to_front() and your having the the hole work done in one function, not a half here and a half there.
void sequence_insert_at(Sequence s, int pos, int item)
{
if(s && pos <= s->length) {
print_sequence(s);
struct node *newnode = malloc(sizeof(struct node));
if (newnode == NULL) {
printf("ERROR! add_to_front ran out of memory!\n");
exit(EXIT_FAILURE);
}
newnode->first = item;
struct node* temp = s->lst;
struct node* prv = NULL;
for(int i = 0; i < pos; i++) {
printf("skip %d\n", temp->first);
prv = temp;
temp = temp->rest;
}
newnode->rest = temp;
if(pos == 0) {
printf("insert as first\n");
s->lst = newnode;
} else {
printf("insert before %d\n", temp->first);
prv->rest = newnode;
}
++s->length;
}
}
and in add_to_front only one statement is needed
void add_to_front(Sequence s, int item) {
sequence_insert_at(s, 0, item);
}
as for inserting at the back of the list
void add_to_back(Sequence s, int item) {
sequence_insert_at(s, s->length, item);
}
A small test with the main function
void print_sequence(Sequence s)
{
struct node* temp = s->lst;
for(int i = 0; i < s->length; temp = temp->rest) {
printf("%d ", temp->first);
i++;
}
printf("\n");
}
int main()
{
Sequence derp = new_sequence();
sequence_insert_at(derp, 0, 14);
add_to_front(derp, 16);
sequence_insert_at(derp, 0, 17);
sequence_insert_at(derp, 2, 15);
add_to_back(derp, 13);
print_sequence(derp);
delete_sequence(derp);
return 0;
}
output is:
17 16 15 14 13
You'll have to go trough the other functions and fix them.
Finally i should note that variable names you have choosen are little bit confusing if not misleading, i would name them this way
typedef struct node {
int data; /* the data that a node holds */
struct node* next; /* the pointer to the next node */
} Node_t;
typedef struct sequence {
struct node* head; /* head or first element of the sequence/list */
int length; /* length is ok but size is better */
} Sequence_t;
I'm trying to implement a hash table as an array of linked lists. Currently I'm trying to have a simple hash table where the key is the index of the array and value is a singly linked list for implementing chaining.
This is the code that I've written so far:
#include<stdio.h>
#include<stdlib.h>
struct Node
{
int value;
struct Node *next;
};
struct Node *hashtable[7];
int empty(int index)
{
if(hashtable[index]==NULL)
return 0;
return 1;
}
void addNode(int frame,struct Node **iter)
{
if(*iter==NULL)
*iter=malloc(sizeof(struct Node));
else
{
while((*iter)->next != NULL)
(*iter)=(*iter)->next;
(*iter)->next=malloc(sizeof(struct Node));
(*iter)=(*iter)->next;
}
(*iter)->value=frame;
(*iter)->next=NULL;
}
void print()
{
int i;
struct Node **iter;
for(i=0;i<7;i++)
{
iter=&hashtable[i];
while(*iter !=NULL)
{
printf("%d%s%d\n",(*iter)->value,"--",i);
(*iter)=(*iter)->next;
}
}
}
int main()
{
int i=0,count=7;
for(i=0;i<7;i++)
hashtable[i]=NULL;
i=empty(1);
printf("%d",i);
do
{
printf("Enter no:\n");
scanf("%d",&i);
struct Node** temp;
temp=&hashtable[i-1%7];
addNode(rand(),temp);
count--;
print();
} while(count > 0);
return 0;
}
When I'm calling print, I can only see one element added to one particular index, which is the last element that was added, what am I doing wrong here?
void add_node(int frame,struct Node **iter)
{
/* find (pointer to) NULL pointer at end of chain */
for ( ; *iter; iter = &(*iter)->next ) {;}
*iter = malloc(sizeof **iter );
(*iter)->value = frame;
(*iter)->next = NULL;
}
I am goofing around with pointers and structures. I want to achieve the following:
(1) define a linked list with a structure (numberRecord)
(2) write a function that fills a linked list with some sample records by going thourgh a loop (fillList)
(3) count the number of elements in the linked list
(4) print the number of elements
I am now so far that the fillList function works well, but I do not succeed in handing over the filled linked list to a pointer in the main(). In the code below, the printList function only displays the single record that was added in main() instead of displaying the list that was created in the function fillList.
#include <stdio.h>
#include <stdlib.h>
typedef struct numberRecord numberRecord;
//linked list
struct numberRecord {
int number;
struct numberRecord *next;
};
//count #records in linked list
int countList(struct numberRecord *record) {
struct numberRecord *index = record;
int i = 0;
if (record == NULL)
return i;
while (index->next != NULL) {
++i;
index = index->next;
}
return i + 1;
}
//print linked list
void printList (struct numberRecord *record) {
struct numberRecord *index = record;
if (index == NULL)
printf("List is empty \n");
while (index != NULL) {
printf("%i \n", index->number);
index = index->next;
}
}
//fill the linked list with some sample records
void fillList(numberRecord *record) {
numberRecord *first, *prev, *new, *buffer;
//as soon as you add more records you get an memory error, static construction
new = (numberRecord *)malloc(100 * sizeof(numberRecord));
new->number = 0;
new->next = NULL;
first = new;
prev = new;
buffer = new;
int i;
for (i = 1; i < 11; i++) {
new++;
new->number = i;
new->next = NULL;
prev->next = new;
prev = prev->next;
}
record = first;
}
int main(void) {
numberRecord *list;
list = malloc(sizeof(numberRecord));
list->number = 1;
list->next = NULL;
fillList(list);
printf("ListCount: %i \n", countList(list));
printList(list);
return 0;
}
SOLUTION
Do read the posts below, they indicated this solution and contain some very insightful remarks about pointers. Below the adapted code that works:
#include <stdio.h>
#include <stdlib.h>
typedef struct numberRecord numberRecord;
//linked list
struct numberRecord {
int number;
struct numberRecord *next;
};
//count #records in linked list
int countList(struct numberRecord *record) {
struct numberRecord *index = record;
int i = 0;
if (record == NULL)
return i;
while (index->next != NULL) {
++i;
index = index->next;
}
return i + 1;
}
//print linked list
void printList (struct numberRecord *record) {
struct numberRecord *index = record;
if (index == NULL)
printf("List is empty \n");
while (index != NULL) {
printf("%i \n", index->number);
index = index->next;
}
}
//fill the linked list with some sample records
numberRecord *fillList() {
numberRecord *firstRec, *prevRec, *newRec;
int i;
for (i = 1; i < 11; i++) {
newRec = malloc(sizeof(numberRecord));
newRec->number = i;
newRec->next = NULL;
//initialize firstRec and prevRec with newRec, firstRec remains head
if (i == 1) {
firstRec = newRec;
prevRec = newRec;
}
prevRec->next = newRec;
prevRec = prevRec->next;
}
return firstRec;
}
int main(void) {
numberRecord *list;
list = fillList();
printf("ListCount: %i \n", countList(list));
printList(list);
return 0;
}
This statement in fillList
record = first;
has no effect on the list variable in main. Pointers are passed by value (like everything else) in C. If you want to update the list variable in main, you'll either have to pass a pointer to it (&list) and modify fillList accordingly, or return a numberRecord* from fillList. (I'd actually go with that second option.)
Here's a (bad) illustration:
When main calls fillList, at the starting point of that function, the pointers are like this:
main memory fillList
list ----> 0x01234 <---- record
A bit later in fillList, you allocate some storage for new (that's actually a bad name, it conflicts with an operator in C++, will get people confused)
main memory fillList
list ----> 0x01234 <---- record
0x03123 <---- new
At the last line of fillList you're left with:
main memory fillList
list ----> 0x01234 ,-- record
0x03123 <---- new
record and list are not the same variable. They start out with the same value, but changing record will not change list. The fact that they are both pointers doesn't make them any different from say ints in this respect.
You can change the thing pointed to by list in fillList, but you can't change what list points to (with your version of the code).
The easiest way for you to get around that is to change fillList like this:
numberRecord *fillList() {
....
return new;
}
And in main, don't allocate list directly, just call fillList() to initialize it.