Im trying create a linked list in an array of nodes. When I try to update the pointer for arrTab->h_table[index] to the address of newNode, The address points to newNodes address. But when I try to add to a list that exists in the array, the pointer always points to NULL instead of the previous value in memory. Basically the arrTab->h_table[index] head of the linked list does not update to the address of newNode.
typedef struct node {
struct node* next;
int hash;
s_type symbol;
} node_t;
struct array {
int cap;
int size;
n_type** h_table;
};
int add_to_array (array* arrTab, const char* name, int address) {
if(s_search(arrTab, name, NULL, NULL) == NULL){
s_type *symbol = (s_type*) calloc(1, sizeof(s_type));
symbol->name = strdup(name);
symbol->addr = addr;
n_type *newNode = (n_type*) calloc(1, sizeof(n_type));
newNode->next = NULL;
newNode->hash = nameHash(name);
newNode->symbol = *symbol;
int index = newNode->hash % arrTab->cap;
if(arrTab->h_table[index] == NULL){
arrTab->h_table[index] = newNode;
} else {
newNode->next = arrTab->h_table[index];
arrTab->h_table[index] = newNode;
}
//
arrTab->size++;
return 1;
}
return 0;
}
struct node* s_search (array* arrTab, const char* name, int* hash, int* index) {
int hashVal = nameHash(name);
hash = &hashVal;
int indexVal = *hash % arrTab->cap;
index = &indexVal;
s_type *symCopy = arrTab;
while (symCopy->h_table[*index] != NULL){
if(*hash == symCopy->h_table[*index]->hash){
return symCopy->h_table[*index];
}
symCopy->h_table[*index] = symCopy->h_table[*index]->next;
}
return NULL;
}
I cannot say for sure why the pointer always points to NULL; there is not enough code. Consider posting an MCVE.
The posted code however presents few problems to address.
First, it leaks memory like there is no tomorrow:
symbol_t *symbol = (symbol_t*) calloc(1, sizeof(symbol_t));
allocates some memory, and
newNode->symbol = *symbol;
copies the contents of that memory to the new location. The memory allocated still exists, and continues to exist after the function returns - but there's no way to get to it. I strongly recommend to not allocate symbol, and work directly with newNode->symbol:
newNode->symbol.name = strdup(name);
newNode->symbol.addr = addr;
The hash and index parameters to symbol_search seem to be planned as an out parameters. In that case, notice that the results of hash = &hashVal; and index = &indexVal; are invisible to the caller. You likely meant *hash = hashVal and *index = indexVal.
The biggest problem comes with sym_table_t *symCopy = symTab;.
symTab is a pointer. It points to an actual symbol table, a big piece of memory. After the assignment, symCopy points to the same piece of memory. Which means that
symCopy->hash_table[*index] = symCopy->hash_table[*index]->next;
modifies that piece of memory. Once the search is completed, the hash_table[index] is not the same as it was before the search. This could be a root of your problem. In any case, consider
node_t * cursor = symTab->hash_table[*index];
and work with this cursor instead.
As a side note, a search condition *hash == symCopy->hash_table[*index]->hash is strange. Every node in a given linked list has the same hash (check how you add them). The very first node would produce a match, even if the names are different.
Related
I have a block of pointers to some structs which I want to handle (i.e. free) separately. As an example below there is an integer double-pointer which should keep other pointers to integer. I then would like to free the second of those integer pointers (in my program based on some filterings and calculations). If I do so however, I should keep track of int-pointers already set free so that when I iterate over the pointers in the double-pointer I do not take the risk of working with them further. Is there a better approach for solving this problem (in ANSI-C) without using other libs (e.g. glib or alike)?
Here is a small simulation of the problem:
#include <stdio.h>
#include <stdlib.h>
int main() {
int **ipp=NULL;
for (int i = 0; i < 3; i++) {
int *ip = malloc(sizeof (int));
printf("%p -> ip %d\n", ip, i);
*ip = i * 10;
if ((ipp = realloc(ipp, sizeof (int *) * (i + 1)))) {
ipp[i] = ip;
}
}
printf("%p -> ipp\n", ipp);
for (int i = 0; i < 3; i++) {
printf("%d. %p %p %d\n", i, ipp + i, *(ipp+i), **(ipp + i));
}
// free the middle integer pointer
free(*(ipp+1));
printf("====\n");
for (int i = 0; i < 3; i++) {
printf("%d. %p %p %d\n", i, ipp + i, *(ipp+i), **(ipp + i));
}
return 0;
}
which prints something like
0x555bcc07f2a0 -> ip 0
0x555bcc07f6f0 -> ip 1
0x555bcc07f710 -> ip 2
0x555bcc07f6d0 -> ipp
0. 0x555bcc07f6d0 0x555bcc07f2a0 0
1. 0x555bcc07f6d8 0x555bcc07f6f0 10
2. 0x555bcc07f6e0 0x555bcc07f710 20
====
0. 0x555bcc07f6d0 0x555bcc07f2a0 0
1. 0x555bcc07f6d8 0x555bcc07f6f0 0
2. 0x555bcc07f6e0 0x555bcc07f710 20
Here I have freed the middle int-pointer. In my actual program I create a new block for an integer double-pointer, iterate over the current one, create new integer pointers and copy the old values into it, realloc the double-pointer block and append the new pointer to it, and at the end free the old block and all it's containing pointers. This is a bit ugly, and resource-consuming if there is a huge amount of data, since I have to iterate over and create and copy all the data twice. Any help is appreciated.
Re:
"This is a bit ugly, and resource-consuming if there is a huge amount of data, since I have to iterate over and create and copy all the data
twice. Any help is appreciated."
First observation: It is not necessary to use realloc() when allocating new memory on a pointer that has already been freed. realloc() is useful when needing to preserve the contents in a particular area of memory, while expanding its size. If that is not a need (which is not in this case) malloc() or calloc() are sufficient. #Marco's suggestion is correct.
Second observation: the following code snippet:
if ((ipp = realloc(ipp, sizeof (int *) * (i + 1)))) {
ipp[i] = ip;
}
is a potential memory leak. If the call to realloc()_ fails, the pointer ipp will be set to null, making the memory location that was previously allocated becomes orphaned, with no way to free it.
Third observation: Your approach is described as needing:
Array of struct
dynamic memory allocation of a 2D array
need to delete elements of 2D array, and ensure they are not referenced once deleted
need to repurpose deleted elements of 2D array
Your initial reaction in comments to considering using an alternative approach notwithstanding, Linked lists are a perfect fit to address the needs stated in your post.
The fundamental element of a Linked List uses a struct
Nodes (elements) of a List are dynamically allocated when created.
Nodes of a List are not accessible to be used once deleted. (No need to track)
Once the need exists, a new node is easily created.
Example struct follows. I like to use a data struct to contain the payload, then use an additional struct as the conveyance, to carry the data when building a Linked List:
typedef struct {//to simulate your struct
int dNum;
char unique_name[30];
double fNum;
} data_s;
typedef struct Node {//conveyance of payload, forward and backward searchable
data_s data;
struct Node *next; // Pointer to next node in DLL
struct Node *prev; // Pointer to previous node in DLL
} list_t;
Creating a list is done by creating a series of nodes as needed during run-time. Typically as records of a database, or lines of a file are read, and the elements of the table record (of element of the line in a file) are read into and instance of the data part of the list_s struct. A function is typically defined to do this, for example
void insert_node(list_s **head, data_s *new)
{
list_s *temp = malloc(sizeof(*temp));
//insert lines to populate
temp.data.dNum = new.dNum;
strcpy(temp.data.unique_name, new.unique_name);
temp.fNum = new.fNum
//arrange list to accomdate new node in new list
temp->next = temp->prev = NULL;
if (!(*head))
(*head) = temp;
else//...or existing list
{
temp->next = *head;
(*head)->prev = temp;
(*head) = temp;
}
}
Deleting a node can be done in multiple ways. It the following example method a unique value of a node member is used, in this case unique_name
void delete_node_by_name(list_s** head_ref, const char *name)
{
BOOL not_found = TRUE;
// if list is empty
if ((*head_ref) == NULL)
return;
list_s *current = *head_ref;
list_s *next = NULL;
// traverse the list up to the end
while (current != NULL && not_found)
{
// if 'name' in node...
if (strcmp(current->data.unique_name, name) == 0)
{
//set loop to exit
not_found = FALSE;
//save current's next node in the pointer 'next' /
next = current->next;
// delete the node pointed to by 'current'
delete_node(head_ref, current);
// reset the pointers
current = next;
}
// increment to next node
else
{
current = current->next;
}
}
}
Where delete_node() is defined as:
void delete_node(list_t **head_ref, list_t *del)
{
// base case
if (*head_ref == NULL || del == NULL)
return;
// If node to be deleted is head node
if (*head_ref == del)
*head_ref = del->next;
// Change next only if node to be deleted is NOT the last node
if (del->next != NULL)
del->next->prev = del->prev;
// Change prev only if node to be deleted is NOT the first node
if (del->prev != NULL)
del->prev->next = del->next;
// Finally, free the memory occupied by del
free(del);
}
This link is an introduction to Linked Lists, and has additional links to other related topic to expand the types of lists that are available.
You could use standard function memmove and then call realloc. For example
Let's assume that currently there are n pointers. Then you can write
free( *(ipp + i ) );
memmove( ipp + i, ipp + i + 1, ( n - i - 1 ) * sizeof( *pp ) );
*( ipp + n - 1 ) = NULL; // if the call of realloc will not be successfull
// then the pointer will be equal to NULL
int **tmp = realloc( ipp, ( n - 1 ) * sizeof( *tmp ) );
if ( tmp != NULL )
{
ipp = tmp;
--n;
}
else
{
// some other actions
}
Here is my implementation of a hashmap in c and its initialization and insert code.
In the hashmap_t structure, I use an array of pointers (table) to nodes which contain the key/value pairs. In hashmap_init, I allocate the desired amount of nodes and loop through the array setting each pointer to NULL.
What I'm confused about is in the hashmap_put function. I find the index of which list the key should be inserted in and that first pointer is referenced by hm->table[i]. For clarity, I want to make sure it's obvious that hm->table[i] is the start of the list so I assign it to hashnode_t *head.
So when inserting the first node (head == NULL), I originally used head = new_node, but none of my inserts worked. It only works when I use hm->table[i] = new_node.
I don't understand why that's the case. head points to the same thing so why does setting head equal to the new_node not work? I'm also confused later in the function when last->next = new_node does work. Last is a pointer just like head but it works there.
Thanks for any clarification.
typedef struct hashnode {
char key[128];
char val[128];
struct hashnode *next;
} hashnode_t;
typedef struct {
int item_count;
int table_size;
hashnode_t **table;
} hashmap_t;
void hashmap_init(hashmap_t *hm, int table_size) {
hm->table_size = table_size;
hm->item_count = 0;
hm->table = malloc(table_size * sizeof(hashnode_t));
for (int i = 0; i < table_size; i++) { // loop through array of pointers to nodes
hm->table[i] = NULL;
}
}
int hashmap_put(hashmap_t *hm, char key[], char val[]) {
hashnode_t *new_node = malloc(sizeof(hashnode_t)); // allocate new node
strcpy(new_node->key, key);
strcpy(new_node->val, val);
new_node->next = NULL;
int i = hashcode(key) % hm->table_size; // index of list hashed to
hashnode_t *head = hm->table[i];
hashnode_t *cur = head;
hashnode_t *last;
if (!head) { // list is empty
new_node->next = head;
hm->table[i] = new_node;
//why does head = new_node not work?
hm->item_count += 1;
return 1;
}
while (cur) { // loop through nodes
if (strcmp(cur->key, key) == 0) {
strcpy(cur->val, val);
free(new_node);
return 0;
}
last = cur; // save pointer to node that points to NULL
cur = cur->next;
}
last->next = new_node;
//why does it work here?
hm->item_count += 1;
return 1;
}
'head' is pointing to a hashnode_t and so is 'hm->table[i]'. So, they are both pointing to the same object. Changing 'head' just makes 'head' point elsewhere. You have not actually assigned a pointer in the hashmap_t to the 'new_node'.
The reason that 'last' works is that you are changing a member variable to a new value. And, since 'last' is pointing to an object already in the hashmap_t, the assignment updates the object pointed to in the hastmap_t. So, an update to 'last->next = new_node' is the same as 'hm->table[x]->next = new_node' ('x' is some arbitrary index).
I don't understand why my program seg faults at this line: if ((**table->table).link == NULL){ I seem to have malloc-ed memory for it, and I tried looking at it with gdb. *table->table was accessible and not NULL, but **table->table was not accessible.
Definition of hash_t:
struct table_s {
struct node_s **table;
size_t bins;
size_t size;
};
typedef struct table_s *hash_t;
void set(hash_t table, char *key, int value){
unsigned int hashnum = hash(key)%table->bins;
printf("%d \n", hashnum);
unsigned int i;
for (i = 0; i<hashnum; i++){
(table->table)++;
}
if (*(table->table) == NULL){
struct node_s n = {key, value, NULL};
struct node_s *np = &n;
*(table->table) = malloc(sizeof(struct node_s));
*(table->table) = np;
}else{
while ( *(table->table) != NULL){
if ((**table->table).link == NULL){
struct node_s n = {key, value, NULL};
struct node_s *np = &n;
(**table->table).link = malloc(sizeof(struct node_s));
(**table->table).link = np;
break;
}else if (strcmp((**table->table).key, key) == 0){
break;
}
*table->table = (**(table->table)).link;
}
if (table->size/table->bins > 1){
rehash(table);
}
}
}
I'm calling set from here:
for (int i = 0; i < trials; i++) {
int sample = rand() % max_num;
sprintf(key, "%d", sample);
set(table, key, sample);
}
Your hashtable works like this: You have bins bins and each bin is a linked list of key / value pairs. All items in a bin share the same hash code modulo the number of bins.
You have probably created the table of bins when you created or initialised the hash table, something like this:
table->table = malloc(table->bins * sizeof(*table->table);
for (size_t i = 0; i < table->bins; i++) table->table[i] = NULL;
Now why does the member table have two stars?
The "inner" star means that the table stores pointers to nodes, not the nodes themselves.
The "outer" start is a handle to allocated memory. If your hash table were of a fixed size, for example always with 256 bins, you could define it as:
struct node_s *table[256];
If you passed this array around, it would become (or "decay into") a pointer to its first element, a struct node_s **, just as the array you got from malloc.
You access the contents of the lĀ“bins via the linked lists and the head of linked list i is table->table[i].
You code has other problems:
What did you want to achieve with (table->table)++? This will make the handle to the allocated memory point not to the first element but tho the next one. After doing that hashnum times, *table->table will now be at the right node, but you will have lost the original handle, which you must retain, because you must pass it to free later when you clean up your hash table. Don't lose the handle to allocated memory! Use another local pointer instead.
You create a local node n and then make a link in your linked list with a pointer to that node. But the node n will be gone after you leave the function and the link will be "stale": It will point to invalid memory. You must also create memory for the node with malloc.
A simple implementation of your has table might be:
void set(hash_t table, char *key, int value)
{
unsigned int hashnum = hash(key) % table->bins;
// create (uninitialised) new node
struct node_s *nnew = malloc(sizeof(*nnew));
// initialise new node, point it to old head
nnew->key = strdup(key);
nnew->value = value;
nnew->link = table->table[hashnum];
// make the new node the new head
table->table[hashnum] = nnew;
}
This makes the new node the head of the linked list. This is not ideal, because if you overwrite items, the new ones will be found (which is good), but the old ones will still be in the table (which isn't good). But that, as they say, is left as an exercise to the reader.
(The strdup function isn't standard, but widely available. It also creates new memory, which you must free later, but it ensures, that the string "lives" (is still valid) after you have ceated the hash table.)
Please not how few stars there are in the code. If there is one star too few, it is in hash_t, where you have typecasted away the pointer nature.
I am reading from txt file into a doubly linked list. The codes can do storing data into Nodes, but when I let it go through the linked list, it got a segmentation fault.
Could you guys please tell what has been wrong with the code, thank you!
This is the data structure:
typedef struct telephoneBookNode {
int id;
char name[NAME_LENGTH];
char telephone[TELEPHONE_LENGTH];
struct telephoneBookNode * previousNode;
struct telephoneBookNode * nextNode;
} TelephoneBookNode;
typedef struct telephoneBookList {
TelephoneBookNode * head;
TelephoneBookNode * tail;
TelephoneBookNode * current;
unsigned size;
} TelephoneBookList;
This is the code to create linked list:
TelephoneBookList * createTelephoneBookList(char entry[]) {
TelephoneBookList* aList = NULL;
TelephoneBookNode* aNode = NULL;
char *tokens;
TelephoneBookNode *(*create)() = createTelephoneBookNode;
aNode = (*create)();
tokens = strtok(entry, ", ");
aNode->id = atoi(tokens);
tokens = strtok(NULL, ", ");
strcpy(aNode->name, tokens);
tokens = strtok(NULL, ", ");
strcpy(aNode->telephone, tokens); //Fine until here
//Do I need this line?
//aList = (TelephoneBookList*) malloc(aList->size + 1) * sizeof aList);
if (aList->head == NULL) {
aNode->nextNode = NULL;
aNode->previousNode = NULL;
aList->current = aNode;
aList->head = aNode;
aList->tail = aNode;
} else {
aList->tail->nextNode = aNode;
aNode->previousNode = aList->tail;
}
return aList;
}
TelephoneBookNode * createTelephoneBookNode() {
TelephoneBookNode* aNode;
aNode = (TelephoneBookNode*) malloc(sizeof *aNode);
return aNode;
}
//Do I need this line?
//aList = (TelephoneBookList*) malloc(aList->size + 1) * sizeof aList);
Yes. Yes you do need that line. Otherwise the next line
if (aList->head == NULL) {
will dereference a null pointer.
Though you already do that in the commented out malloc call, dereference a null pointer, with aList->size + 1.
The correct line should be
aList = malloc(sizeof *aList);
And since you create the list from scratch in the function, there is no need to check if it is empty or not, it will always be empty. More importantly, the malloc call will not initialize the memory it allocates, so using that memory (for example in an expression like aList->head == NULL) will lead to undefined behavior.
Allocate the list structure. And then initialize it as if it was empty. And don't forget to initialize the size member as well.
Your createTelephoneBookNode function does not initialize the node that it created. malloc() assigns it a memory block that's probably not been initialized with zeros, and as a result, the nextNode and previousNode pointers contain garbage. Either set them both to NULL, or allocate your memory with calloc().
I trying to write a queue(String Version) program in C by using linked lists.
Here is the structure:
struct strqueue;
typedef struct strqueue *StrQueue;
struct node {
char *item;
struct node *next;
};
struct strqueue {
struct node *front;//first element
struct node *back;//last element in the list
int length;
};
I creates a new StrQueue first
StrQueue create_StrQueue(void) {
StrQueue q = malloc(sizeof (struct strqueue));
q->front = NULL;
q->back = NULL;
q->length = 0;
return q;
}
makes a copy of str and places it at the end of the queue
void push(StrQueue sq, const char *str) {
struct node *new = malloc(sizeof(struct node));
new->item = NULL;
strcpy(new->item,str);//invalid write size of 1 ?
new->next = NULL;
if (sq->length == 0) {
sq->front = new;
sq->back = new;
} else {
sq->back->next = new;
sq->back = new;
}
sq->length++;
}
frees the node at the front of the sq and returns the string that was first in the queue
char *pop(StrQueue sq) {
if (sq->length == 0) {
return NULL;
}
struct node *i = sq->front;
char *new = sq->front->item;
sq->front = i->next;
sq->length --;
free(sq->front);
return new;
}
I got invalid write size of 1 at strcpy(new->item,str); I dont understand why I got this error.
Can anyone tell me why and tell me how should I fix it? Thanks in advance.
Okay, first things first, in the answer below I am NOT fixing your doubly linked list concepts, I am just showing you how you should fix the code above within the scope of your question. You may want to look into how doubly linked lists are done.
In:
void push(StrQueue sq, const char *str) {
struct node *new = malloc(sizeof(struct node));
new->item = NULL;
The next statement is wrong:
strcpy(new->item,str);
There are two ways you can solve it:
Make sure that *str is a valid pointer outside of the list management context while the list is being used.
Let the list manage the string allocation (and possibly deallocation).
is the quick and dirty method, it's easier to debug later but larger codebase makes it cumbersome.
cleaner looking code, but requires initial setup discipline, you should create object (string) management routines in addition to list management routines. can be cumbersome in its own right.
CASE 1: const char *str is guaranteed to be valid for life of StrQueue (this is what you are looking for really)
It should be:
new->item = str;
Here we assume str was a dynamic string allocated elsewhere
Now, in pop when you pop off the string you are okay. because the pointer you are returning is still valid (you are guaranteeing it elsewhere)
CASE 2: const char *str is not guaranteed to be valid for life of StrQueue
Then use:
new->item = strdup(str);
Now, in pop when you pop off the string you can either
de-allocate the strdup and not return anything, (not quite the same things as you did)
pass a container pointer to pop where contents of item are copied (clean)
return the popped off pointer, but you must deallocate it separately when you are done with it (ugly)
Which would make your pop function one of the following:
Case 2.1:
void pop(StrQueue sq) {
if (sq->length == 0) {
return NULL;
}
struct node *node = sq->front;
sq->front = node->next;
sq->length--;
free(node->item);
free(node);
}
Case 2.2:
char *pop(StrQueue sq, char *here) {
if (sq->length == 0) {
return NULL;
}
struct node *node = sq->front;
sq->front = node->next;
sq->length--;
strcpy(here, node->item);
free(node->item);
free(node);
}
Case 2.3:
char *pop(StrQueue sq) {
char *dangling_item = NULL;
if (sq->length == 0) {
return NULL;
}
struct node *node = sq->front;
sq->front = node->next;
sq->length--;
dangling_item = node->item;
free(node);
return dangling_item;
}
I got invalid write size of 1 at strcpy(new->item,str); I dont understand why I got this error. Can anyone tell me why and tell me how should I fix it?
Why:
This code:
new->item = NULL;
strcpy(new->item,str);//invalid write size of 1 ?
You're not suppose to pass a null pointer to the first argument, it should be a pointer to allocated memory. The reason why you're getting this error message, I can imagine, is because the implementation of strcpy probably looks like this:
for (int i = 0; str2[i]; i++) str1[i] = str2[i];
And in the first iteration of the for loop, it writes to address 0 (a read-only section of memory) - this gives you the invalid write of size 1. I'm not sure, however, why you are only getting a size of 1, though (I would imagine it would be the entire size of the string). This could be because either a) str is only of size 1 or b) because the signal, SIGSEGV stops the program.
How to fix:
Allocate space for new->item before calling strcpy, like this:
new->item = malloc (strlen (str) + 1); // + 1 for null-terminating character
But you could probably include some error checking, like this:
int len = strlen (str) + 1;
if (len){
new->item = malloc (len);
if (!new->item){
return;
}
}