I tried to implement a binary search tree for the purpose of (re-)learning C. The problem is that the this current = new; does not work as desired because the tree.root is still a null pointer after adding two nodes. What's wrong with that?
#include <stdlib.h>
#include <stdio.h>
typedef struct BinaryNode {
int key;
double value;
struct BinaryNode *left;
struct BinaryNode *right;
} BinaryNode;
typedef struct BinaryTree {
struct BinaryNode *root;
} BinaryTree;
static void binary_tree_insert_recursive(BinaryNode *current, BinaryNode *new) {
if (current == NULL || current->key == new->key) {
current = new;
} else if (current->key > new->key) {
binary_tree_insert_recursive(current->left, new);
} else if (current->key < new->key) {
binary_tree_insert_recursive(current->right, new);
}
}
void binary_tree_insert(BinaryTree *tree, int key, double value) {
BinaryNode *new = (BinaryNode *) malloc(sizeof(BinaryNode));
new->key = key;
new->value = value;
binary_tree_insert_recursive(tree->root, new);
}
int main(void) {
BinaryTree tree;
binary_tree_insert(&tree, 5, 123);
binary_tree_insert(&tree, 10, 123);
printf("%p\n", tree.root);
return 0;
}
Thank you!
I believe the problem with current = new; is that you are changing your local copy of current. After the function is done, this modification is not visible.
I suspect you want something like:
static void binary_tree_insert_recursive(BinaryNode **current, BinaryNode **new)
{
if (*current == NULL || (*current)->key == (*new)->key) {
*current = *new;
/* ... */
Well explained in the C FAQ.
current is a pointer to a node. When you pass it to binary_tree_insert_recursive from binary_tree_insert the value of the pointer is passed. So although it is changed inside the called function, the change is not reflected in the calling function. You need to modify the function to take the address of the pointer you wish to change:
static void binary_tree_insert_recursive(BinaryNode **current, BinaryNode *new)
{
if (*current == NULL || (*current)->key == new->key) {
*current = new;
all that current = new does is make it so that the variable current points at the thing that new is pointing at. No copying takes place, and the function has no effect on that codepath.
new is a keyword. Choose a different variable name.
Related
I'm currently doing an assignment for uni and I need to find the sum of a graph.
To do this I believe I need a linked list that I can use to remember which nodes have been visited. I have the linkedlist working correctly but I can't get a contains function to work. This is the code I have:
struct listnode
{
struct N *val;
struct listnode *next;
};
int contains(struct listnode *head,struct N* value)
{
struct listnode *current = head;
while (current)
{
if ((current -> val) == value)
{
return 1;
}
current = current -> next;
}
return 0;
}
note: N is a node of the graph.
Can anyone see any problems with what I'm doing?
EDIT: contains function should return 1 when N *value is in the list, 0 otherwise
EDIT2:
I have a push function:
void push(struct listnode *head,struct N *value)
{
if (head)
{
struct listnode *current = head;
while (current->next)
{
current = current -> next;
}
current->next = malloc(sizeof(struct listnode*));
current->next->val = value;
current->next->next = NULL;
}
else
{
head = malloc(sizeof(struct listnode*));
if (head)
{
head -> val = value;
head -> next = NULL;
}
else
{
printf("error");
exit(0);
}
}
}
and I want the following line to return 1:
contains(push(visited,p),p);
where p is a pointer to a struct N and visited is my global linked list
EDIT3:
this is my final sum function that I believe should work, but doesnt because of contains.
long sum(struct N *p)
{
if (p)
{
if (contains(visited,p) == 0) //if p hasnt been visited
{
push(visited,p); //make it visited
return (p -> data) + sum(p -> x) + sum(p -> y) + sum(p -> z);
}
else
{
return 0;
}
}
else
{
return 0;
}
}
Your contains function appears to be fine. The issue is that you are always passing a NULL list to it, which is caused by a faulty push function. You need a return in push, or to pass in a pointer with one more level of indirection, so you can assign to head outside of push. One more possible improvement is to notice that no matter what you pass in, the malloc and initialization of a new node is actually the same.
Finally, the main issue, that is really the most likely to cause a segfault is the fact that you are allocating enough space for a pointer to a node, not for the node itself.
Here is an example:
#ifdef BY_INDIRECTION
#define RET_TYPE void
#define IN_TYPE struct listnode **
#else
#define RET_TYPE struct listnode *
#define IN_TYPE struct listnode *
#endif
RET_TYPE push(IN_TYPE head, struct N *value)
{
struct listnode *current, **next;
if(head)
{
for(current = head; current->next; current = current->next) ;
next = &(current->next);
}
else
{
#ifdef BY_INDIRECTION
next = head;
#else
next = &head;
#endif
}
*next = malloc(sizeof(struct listnode));
if(!*next) {
printf("error");
exit(0);
}
(*next)->val = value;
(*next)->next = NULL;
#ifndef BY_INDIRECTION
return head
#endif
}
I have included both suggestions here. If you want to read the one where we use indirection (pass in a listnode ** and have void return), choose the path where BY_INDIRECTION is defined. If you want to have head returned (and pass in just a regular listnode *) read the path where BY_INDIRECTION is not defined.
The latter approach has a return value, so it can be used to write a shortened form like if(contains(push(head, value), value)) { ... }. The former approach does not, so you would have to do
push(&head, value);
if(contains(head, value)) { ... }
I would recommend using the indirect approach regardless because there are very few instances that you would want to check for containment after putting in a value.
This comparison:
if ((current -> val) == value)
it's comparing pointers. If you call your contains() function this way...
...
struct N val_to_find;
...
result = contains (list, &val_to_find);
You will never find the value, even if the contents of val_to_find are the same as the contents of any struct whose pointer is stored in the list.
If your intention for contains() is to find nodes that have the same data, and not just the same pointers, I'd suggest you something like this:
if (struct_n_comparing_function (current -> val, value) == EQUAL) ...
Where struct_n_comparing_function should have the following prototype:
int struct_n_comparing_function (struct N *a, struct N *b);
which compares the contents of the two structs pointed by a and b and return EQUAL if all the fields of the struct pointed by a have the same value as the fields of struct pointed by b.
I know how pointers works.
I done similar problem with this way
deleteNode(struct node *head_ref, int key);
which is working and # here http://quiz.geeksforgeeks.org/linked-list-set-3-deleting-node/ they have used
deleteNode(struct node **head_ref, int key);
which also correct but is there reason to do so , will 1st one fails in any condition or is it bad way etc.
struct linked_list *deleteNode(struct linked_list *head, int key )
{
struct linked_list *prevNode,*current,*temp;
if( head==NULL)
return head;
if(head->data==key)
{
if(head->next==NULL)
{ free(head);
return NULL;
}
else
temp=head->next;
free(head);
return temp;
}
prevNode= head;
current=head->next;
printf("\n %d\n",(current->data));
while((current!=NULL) && (current->data!=key))
{ printf("\n here");
prevNode= current;
current=current->next;
}
if(current==NULL){
printf("\n element not present in list !\n");
return head;
}
if(current->next==NULL)
prevNode->next=NULL;
else
prevNode->next=current->next;
free(current);
return head;
}
head=deleteNode(head,key);
If you need to delete the head node, the first function won't work because you can't change the head node. The second function takes the address of the head node so it can be changed if need be.
The deleteNode function in the link contains the following:
struct node* temp = *head_ref, *prev;
// If head node itself holds the key to be deleted
if (temp != NULL && temp->data == key)
{
*head_ref = temp->next; // Changed head
free(temp); // free old head
return;
}
You can see here that it dereferences head_ref to change what it points to.
Let's forget the linked list and just think of updating a variable. There are two, equally valid ways to do it:
// 1. pass back
int update_int1(int val) {
return val + 1;
}
void caller1() {
int var = 1;
var = update_int1(var);
}
// 2. write back
void update_int2(int *val) {
*val += 1;
}
void caller2() {
int var = 1;
update_int2(&var);
}
This is easy to understand, so let's do the same thing with a pointer:
// 1. pass back
char *update_ptr1(char *ptr) {
return ptr + 1;
}
void caller1() {
char *ptr = malloc(10);
ptr = update_ptr1(ptr);
}
// 2. write back
void update_ptr2(char **ptr) {
*ptr += 1;
}
void caller2() {
char *ptr = malloc(10);
update_ptr2(&ptr);
}
It works exactly the same as for int! The key is there's always one more star if you want to write back, not pass back.
Which pattern you choose is up to you. The write-back approach is popular for linked lists.
When you write *b==>access contents of address contained in b.
When you write **c==>Access contents of contents of address contained in c.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
typedef struct node_{
int val;
struct node_ *left;
struct node_ *right;
}node;
node* insert(node* root,int val);
void inorder(node* root);
int main(void)
{
int i;
int item;
node* root = NULL;
srand(time(NULL));
for( i = 0; i < 10; i++)
{
item = rand()%15;
insert(root,item);
}
inorder(root);
return 0;
}
node* insert(node* root,int val)
{
if(root == NULL)
{
root = malloc(sizeof(node));
if(root!= NULL)
{
(root)->val = val;
(root)->left = NULL;
(root)->right = NULL;
}
else
printf("%d not inserted. No memory available.\n",val);
}
else
{
if(val < (root)->val)
{
insert((root->left),val);
}
if(val>root->val)
{
insert(((root)->right),val);
}
}
}
void inorder(node* root)
{
printf("%p",root);
if(root != NULL)
{
inorder(root->left);
printf("%3d",root->val);
inorder(root->right);
}
}
I am trying to create a binary tree and print out the values in order. However when I run this code the printf of the address prints out nil obviously meaning that my tree is empty so the printf and recursion below does not run. I cannot figure out where I went wrong, any suggestions or answers would be appreciated because I can't figure out why the root would be null after calling all of those inserts in main.
You pass root as a parameter to insert() (which says it is going to return something but doesn't). Inside insert you malloc your node and assign it to the local variable root. Nothing you ever do makes it out of the insert function.
Try returning something from insert, or using a global root.
As #JoshuaByer hints in the comments below, another approach is to make your insert method "pass by reference" so it can effectively modify what was passed to it.
void insert(node** rootp,int val)
{
if(*rootp == NULL)
{
*rootp = malloc(sizeof(node));
}
/* and so on */
If you don't understand what this is saying, google "Pass by reference in C" and I'm positive you'll get some good information.
In main() after declaring and initializing root (node* root = NULL;) you're never assigning it. In order to fix you should probably change the lin insert(root,item); to root = insert(root,item);.
Also note that although insert is defined as returning node * it does not return any value.
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 am working for a C exam and while trying to insert an element to a linked list, I am encountering with a runtime problem. My only purpose is adding 4 elements to list and then printing the list. However, it gives an error. I already looked some insertion codes and my code seems right. Can't see the error. Any assistance would be appreciated.
#include <stdio.h>
#include <stdlib.h>
struct ders{
char kod;
struct ders *next;
}*header;
typedef struct ders Ders;
void add(Ders*,Ders*);
void print(Ders*);
int main(void)
{
header = NULL;
Ders *node = NULL;
int i = 0;
char c;
while(i<4)
{
scanf("%c",&c);
node = (Ders*)malloc(sizeof(Ders));
node->kod = c;
node->next = NULL;
add(header,node );
i++;
}
print(header);
return 0;
}
void add(Ders *header, Ders *node)
{
if(header == NULL){
header = node;
header->next = NULL; }
else{
node->next = header;
header = node;
}
}
void print(Ders *header)
{
Ders *gecici = header;
while(gecici != NULL){
printf("%c\n",gecici->kod);
gecici = gecici->next;
}
}
As nihirus stated,
"The pointer is passed by value. Thus you can change the memory it points but you can't change the actual pointer, i.e. make it point to something else."
Your modification resulted in error *header is not member of struct
because
->
has a higher precedence than
*
Try using
(*header)->next = NULL
instead.
C operator precedence:
http://www.difranco.net/compsci/C_Operator_Precedence_Table.htm