I want to add a Newnode to the end of a list but it crashes and goes into a endless loop.
I am attaching the function:
hope to your help!!
void AddProduct(products **head,products *newProduct)
{
products* current=*head;
if(current == NULL)
{
(*head) =(products *)malloc(1*sizeof(products));
(*head) = newProduct;
current=*head;
return;
}
while(current->nextProduct!=NULL)
{
current=current->nextProduct;
}
//Attaching the new product to the list
current->nextProduct=newProduct;
newProduct->prevous=current;
//SortList(head);
}
When creating a new element, you're both accepting it as an argument, and allocating it yourself inside the function. You're overwriting the newly allocated pointer with the one passed in, it doesn't make any sense.
Also, you're not setting the nextProduct field of the new element.
Finally, you're using too many casts and parentheses.
Related
This is my function for add an element at the end of my lista but I can't find a way for fix the loop in the while, can u give me some tips/rules for make this function work?
void insCoda(t_lista *l, TipoElemLista elem){
t_lista ultimo;
t_lista temp;
temp=(node *)malloc(sizeof(node));
temp->contenuto=elem;
temp->next= NULL;
if(*l==NULL)
{
*l=temp;
printf("Dentro if");
}else{
ultimo=*l;
while(ultimo->next!=NULL)
{
ultimo=ultimo->next;
ultimo->next=temp;
}
}
}
This is thoroughly broken:
ultimo=*l;
while(ultimo->next!=NULL)
{
ultimo=ultimo->next;
ultimo->next=temp;
}
On entry, you set ultimo to (presumably) the head of the list. Then you advance past it to the next node (ultimo=ultimo->next), and immediately set the next pointer of that node to your newly allocated node (ultimo->next=temp). Except oops, your very next action is to test if the thing you just set is NULL or not (and it isn't, unless malloc failed). So you process your new node, and set its next to itself. And now you're in an infinite loop. If you don't enter the loop (because your head is the only node, so the loop condition fails immediately), you never insert the new node at all (which is nice, because this is saving you from the infinite loop).
A hint: Don't set next inside the loop. While I haven't tested, simply moving the set outside the loop should work:
ultimo=*l;
while(ultimo->next!=NULL)
{
ultimo=ultimo->next;
}
ultimo->next=temp;
so now you traverse to the final node, then make your new node the final node.
Now, I understand that code below works only for root and its children, but I don't know how to expand it. Every node must have children before passing on "grandchildren". Thank you.
void insert_node(IndexTree **root, Node *node) {
IndexTree *temp = (IndexTree*)malloc(sizeof(IndexTree));
memcpy(&temp->value.cs, node, sizeof(Node));
temp->left = NULL;
temp->right = NULL;
temp->tip=1;
if ((*root) == NULL) {
*root = temp;
(*root)->left = NULL;
(*root)->right = NULL;
}
else {
while (1) {
if ((*root)->right == NULL) {
(*root)->right = temp;
break;
}
else if ((*root)->left == NULL) {
(*root)->left = temp;
break;
}
}
}
Use recursive functions.
Trees are recursive data types (https://en.wikipedia.org/wiki/Recursive_data_type). In them, every node is the root of its own tree. Trying to work with them using nested ifs and whiles is simply going to limit you on the depth of the tree.
Consider the following function: void print_tree(IndexTree* root).
An implementation that goes over all values of the trees does the following:
void print_tree(IndexTree* root)
{
if (root == NULL) return; // do NOT try to display a non-existent tree
print_tree(root->right);
printf("%d\n", root->tip);
print_tree(root->left);
}
The function calls itself, which is a perfectly legal move, in order to ensure that you can parse an (almost) arbitrarily deep tree. Beware, however, of infinite recursion! If your tree has cycles (and is therefore not a tree), or if you forget to include an exit condition, you will get an error called... a Stack Overflow! Your program will effectively try to add infinite function calls on the stack, which your OS will almost certainly dislike.
As for inserting, the solution itself is similar to that of printing the tree:
void insert_value(IndexTree* root, int v)
{
if (v > root->tip) {
if (root->right != NULL) {
insert_value(root->right, v);
} else {
// create node at root->right
}
} else {
// same as above except with root->left
}
}
It may be an interesting programming question to create a Complete Binary Tree using linked representation. Here Linked mean a non-array representation where left and right pointers(or references) are used to refer left and right children respectively. How to write an insert function that always adds a new node in the last level and at the leftmost available position?
To create a linked complete binary tree, we need to keep track of the nodes in a level order fashion such that the next node to be inserted lies in the leftmost position. A queue data structure can be used to keep track of the inserted nodes.
Following are steps to insert a new node in Complete Binary Tree. (Right sckewed)
1. If the tree is empty, initialize the root with new node.
2. Else, get the front node of the queue.
……. if the right child of this front node doesn’t exist, set the right child as the new node. //as per your case
…….else If the left child of this front node doesn’t exist, set the left child as the new node.
3. If the front node has both the left child and right child, Dequeue() it.
4. Enqueue() the new node.
I am working on a C program which has a linked list. I need to remove the last element from the linked list and it is mostly working except when it hits particular part of my code it then has a segmentation fault.
The code that I have is as follows:
int clearOutboundLegFromList(callLogSearchOutboundStruct ** outboundLeg, int dataCol, int rowTargets)
{
//callLogSearchOutboundStruct *currentStruct = *outboundLeg;
//callLogSearchOutboundStruct *temp;
if (*outboundLeg == NULL)
{
return 0;
}
SL_DebugAll(DBG_ALWAYS, "DEBUG: Clearing outbound legs: DataCol: %i RowTargets: %i",
dataCol, rowTargets);
callLogSearchOutboundStruct *legToRemove = NULL;
callLogSearchOutboundStruct *last = NULL;
legToRemove = *outboundLeg;
while (legToRemove->nextLeg != NULL)
{
last = legToRemove;
legToRemove = legToRemove->nextLeg;
}
if (legToRemove->target != NULL)
{
free(legToRemove->target);
legToRemove->target = NULL;
}
if (legToRemove->cleardownCause)
{
free(legToRemove->cleardownCause);
legToRemove->cleardownCause = NULL;
}
free(legToRemove);
if (last != NULL)
{
last->nextLeg = NULL;
}
legToRemove = NULL;
}
It crashes on the line of free(legToRemove->target);.
In the core dump I have the following:
Program terminated with signal 11, Segmentation fault.
#0 0x00b01336 in _int_free () from /lib/libc.so.6
Missing separate debuginfos, use: debuginfo-install cyrus-sasl-lib-2.1.23-13.el6_3.1.i686 glibc-2.12-1.132.el6_5.2.i686 keyutils-libs-1.4-4.el6.i686 krb5-libs-1.10.3-15.el6_5.1.i686 libcom_err-1.41.12-18.el6.i686 libcurl-7.19.7-37.el6_5.3.i686 libidn-1.18-2.el6.i686 libselinux-2.0.94-5.3.el6_4.1.i686 libssh2-1.4.2-1.el6.i686 mysql-libs-5.1.73-3.el6_5.i686 nspr-4.9.2-1.el6.i686 nss-3.14.0.0-12.el6.i686 nss-softokn-freebl-3.12.9-11.el6.i686 nss-util-3.14.0.0-2.el6.i686 openldap-2.4.23-31.el6.i686 openssl-1.0.1e-16.el6_5.14.i686 zlib-1.2.3-29.el6.i686
(gdb) bt
#0 0x00b01336 in _int_free () from /lib/libc.so.6
#1 0x0805cd0b in clearOutboundLegFromList (outboundLeg=0xb5de7984, dataCol=9, rowTargets=11) at performreport.c:6731
#2 0x08058f33 in processDrilldownData (reportParameterArray=..., csvFile=0x8e3fc78, HandleDB=0xbfca7a14, resultReport=0x8e457a8,
If I print from the core dump legToRemove->target gdb outputs the following:
$1 = 0x99235d8 ""
Now that looks like its a properly allocated memory space, it just contains an empty string so I don't understand why this would cause a segfault.
You don't show how your struct looks like or how you add legs to your linked list, but you have an error in your removal function that occurs if you remove the last node: In that case, your list head should be set to NULL.
This special case is the reason to pass the list head as pointer to pointer to leg: The function must be able to update the head when the first node is removed. If you don't do that, the value of the head in the calling function will be the same and it will refer to memory that you have just freed. It is illegal to access such memory.
So, an updated version of your code could look like this:
void clearOutboundLegFromList(callLogSearchOutboundStruct **outboundLeg)
{
callLogSearchOutboundStruct *last = NULL;
legToRemove = *outboundLeg;
if (legToRemove == NULL) return;
while (legToRemove->nextLeg) {
last = legToRemove;
legToRemove = legToRemove->nextLeg;
}
free(legToRemove->target);
free(legToRemove->cleardownCause);
free(legToRemove);
if (last) {
last->nextLeg = NULL;
} else {
*outboundLeg = NULL;
}
}
You need the explicit assignment at the end, because once you have initialised legToRemove, you are operating only with that local pointer.
If you are feeling more confident with double indirections via pointers to pointers, you could iterate to the end without local variabes:
void clearOutboundLegFromList(callLogSearchOutboundStruct **outboundLeg)
{
if (*outboundLeg == NULL) return;
while (*outboundLeg) {
outboundLeg = &(*outboundLeg)->nextLeg;
}
free((*outboundLeg)->target);
free((*outboundLeg)->cleardownCause);
free(*outboundLeg);
*outboundLeg = NULL;
}
This will update the head pointer automatically when the first element is removed. The idea here ist that outboundLeg points to the head node at the beginning and to the previous node's nextLeg pointer on subsequent iterations. The additional indirection via (*outboundLeg) is more or less the same as accessing a node via the nextLeg member, except for the first node, in which you access the pointer through the head node pointer.
(Distraction: Your code is overly cautious when freeing the member pointers. It is legal to free a null pointer; this doesn't do anything, but means that you don't have to check for NULL in client code. Such a check might still be good practice, because many functions won't take null pointers. Setting the member pointers to NULL is a good idea if these pointers were still around for some time. But you are going to free the containing struct anyway soon. Setting the pointers to NULL is a bit like cleaning the bathroom just before you tear down the house. Setting legToRemove to NULL at the end of the function doesn't do anything: The pointer will go out of scope anyway. That's just an aside and retionale for my shorter code. Your checks aren't wrong and it is better to be cautious.)
I'm now implementing Barnes-Hut Algorithms for simulating N-body problem. I only want to ask about the building-tree part.
There are two functions I made to build the tree for it.
I recursively build the tree, and print the data of each node while building and everything seems correct, but when the program is back to the main function only the root of the tree and the child of the root stores the value. Other nodes' values are not stored, which is weird since I printed them during the recursion and they should have been stored.
Here's some part of the code with modification, which I thought where the problem might be in:
#include<...>
typedef struct node{
int data;
struct node *child1,*child2;
}Node;
Node root; // a global variable
int main(){
.
set_root_and_build(); // is called not only once cuz it's actually in a loop
traverse(&root);
.
}
Here's the function set_root_and_build():
I've set the child pointers to NULL, but didn't show it at first.
void set_root_and_build(){
root.data = ...;
..// set child1 and child2 =NULL;
build(&root,...); // ... part are values of data for it's child
}
And build:
void build(Node *n,...){
Node *new1, *new2 ;
new1 = (Node*)malloc(sizeof(Node));
new2 = (Node*)malloc(sizeof(Node));
... // (set data of new1 and new2 **,also their children are set NULL**)
if(some condition holds for child1){ // else no link, so n->child1 should be NULL
build(new1,...);
n->child1 = new1;
//for debugging, print data of n->child1 & and->child2
}
if(some condition holds for child2){ // else no link, so n->child2 should be NULL
build(new2,...);
n->child1 = new2;
//for debugging, print data of n->child1 & and->child2
}
}
Nodes in the tree may have 1~2 children, not all have 2 children here.
The program prints out the correct data when it's in build() function recursion, but when it is back to main function and calls traverse(), it fails due to a segmentation fault.
I tried to print everything in traverse() and found that only the root, and root.child1, root.child2 stores the value just as what I've mentioned.
Since I have to called build() several times, and even in parallel, new1 and new2 can't be defined as global variables. (but I don't think they cause the problem here).
Does anyone know where it goes wrong?
The traverse part with debugging info:
void traverse(Node n){
...//print out data of n
if(n.child1!=NULL)
traverse(*(n.child1))
...//same for child2
}
You may not be properly setting the children of n when the condition does not hold. You might want this instead:
void set_root_and_build()
{
root.data = ...;
build(&root,...); // ... part are values of data for it's child
}
void build(Node *n,...)
{
n->child1 = n->child2 = NULL;
Node *new1, *new2;
new1 = (Node*) malloc(sizeof(Node));
new2 = (Node*) malloc(sizeof(Node));
// set data of new1 and new2 somehow (read from stdin?)
if (some condition holds for new1)
{
n->child1 = new1;
build(n->child1,...);
//for debugging, print data of n->child1
}
else
free(new1); // or whatever else you need to do to reclaim new1
if (some condition holds for new2)
{
n->child2 = new2;
build(n->child2,...);
//for debugging, print data of n->child2
}
else
free(new2); // or whatever else you need to do to reclaim new2
}
Of course, you should be checking the return values of malloc() and handling errors too.
Also, your traversal is a bit strange as it recurses by copy rather than reference. Do you have a good reason for doing that? If not, then maybe you want:
void traverse(Node *n)
{
...//print out data of n
if (n->child1 != NULL)
traverse(n->child1)
...//same for child2
}
The problem in your tree traversal is that you certainly process the tree until you find a node pointer which is NULL.
Unfortunately when you create the nodes, these are not initialized neither with malloc() nor with new (it would be initialized with calloc() but this practice in cpp code is as bad as malloc()). So your traversal continues to loop/recurse in the neverland of random pointers.
I propose you to take benefit of cpp and change slightly your structure to:
struct Node { // that's C++: no need for typedef
int data;
struct node *child1,*child2;
Node() : data(0), child1(nullptr), child2(nullptr) {} // Makes sure that every created are first initalized
};
And later get rid of your old mallocs. And structure the code to avoid unnecessary allocations:
if(some condition holds for child1){ // else no link, so n->child1 should be NULL
new1=new Node; // if you init it here, no need to free in an else !!
build(new1,...);
n->child1 = new1;
...
}
if (... child2) { ... }
Be aware however that poitners allocated with new should be released with delete and note with free().
Edit: There is a mismatch in your code snippet:
traverse(&root); // you send here a Node*
void traverse(Node n){ // but your function defines an argument by value !
...
}
Check that you didn't overllok some warnings from the compiler, and that you have no abusive cast in your code.
I wanna make pop function to delete the node and subtree of the node. Here is my code
void pop(struct data *node,int num)
{
if(node)
{
if(node->num==num)
{
pop(node->left,num);
pop(node->right,num);
free(node);
node=NULL;
}
else
{
if(num> node->num)
pop(node->right,num);
else if (num< node->num)
pop(node->left,num);
}
}
}
void pre(struct data *node)
{
if(node)
{
printf("%d ",node->num);
pre(node->left);
pre(node->right);
}
}
void main()
{
push(&root,37);
push(&root,20);
push(&root,45);
push(&root,5);
push(&root,15);
push(&root,40);
push(&root,50);
pre(root);
pop(root,5);
pre(root);
getchar();
}
Pre function works well before I use pop. But after I used the pop function, it's break. Could anyone knows where's the mistake?
In pop, you're doing: node=NULL; -- but this only affects the copy of the pointer that was passed to the function, not the pointer in the original tree. Your tree retains the pointer to the data you've now freed. The next time you do much with the tree, you try to dereference that pointer, and things fall down and go boom (at least you hope they do -- even worse, sometimes they might seem to work).
One way to fix this is to pass a double pointer to pop:
void pop(struct data **node, int num) {
if ((*node)->num == num)
// ...
free(*node);
*node = NULL;
}
}
Now you're changing the pointer in the tree instead of changing the copy of it your function received.
This still won't work quite right though -- you're depending on pop(child, num); to destroy the sub-trees of the current node, but unless their num is set to the same value, they won't delete anything, just travel down the tree looking for a node with a matching num.
You probably want one function to walk the tree finding the node you care about, then a second one that walks the tree starting from a designated node, and (unconditionally) destroys that node and its sub-trees.
Well your pop function should be like this:
struct data* pop(struct data *node,int num)
{
struct data* temp=null;
if(node)
{
if(node->num==num)
{
if(node->left)
pop(node->left,node->left->num);
if(node->right)
pop(node->right,node->right->num);
free(node);
}
else
{
if(num> node->num)
temp=pop(node->right,num);
else if (num< node->num)
temp=pop(node->left,num);
if(node->right==temp)
node->right=null;
else if(node->left==temp)
node->left=null;
return temp;
}
}
return node;
}
This will work as far as you have the logic to nullify the root of the tree from where it is being called, if the desired node tuned out to be root of the tree.