I tried to make a simpler example of my problem, I hope the code doesn't look too weird. I am trying to make a sorted list. If the input is for ex wrt, er, ff, fr, when I try to print the list it says fr fr fr fr and I can't understand why.
Unfortunately I can't post the original code, the struct tit doesn't make much sense in this example but I don't think that's the problem.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct nodo *link;
typedef struct lista *TLIST;
typedef struct tit *TITOLO;
TLIST LISTinit();
link NEW(TITOLO val, link next);
void LISTinsert(TLIST l, TITOLO val);
void ListPrint(TLIST list);
TLIST caricaFile();
struct tit {
char nome[20];
};
struct lista {
link head;
int n;
};
struct nodo {
TITOLO val;
link next;
};
int main() {
TLIST lista;
lista = caricaFile();
ListPrint(lista);
return 0;
}
TLIST LISTinit() {
TLIST list = malloc(sizeof *list);
list->head = NULL;
list->n = 0;
return list;
}
link NEW(TITOLO val, link next) {
link x = malloc(sizeof *x);
x->next = next;
x->val = val;
return x;
}
void LISTinsert(TLIST l, TITOLO val) {
link x, p;
l->n++;
if (l->head == NULL || (strcmp(val->nome, l->head->val->nome) < 0)) {
l->head = NEW(val, l->head);
return;
}
for (x = l->head->next, p = l->head;
x != NULL && (strcmp(val->nome, x->val->nome) > 0);
p = x, x = x->next);
p->next = NEW(val, x);
}
void ListPrint(TLIST list) {
link x;
for (x = list->head; x != NULL; x = x->next)
printf("%s ", x->val->nome);
printf("\n");
}
TLIST caricaFile() {
TLIST list;
list = LISTinit();
TITOLO titolo = malloc(sizeof *titolo);
int i;
for (i = 0; i < 4; i++) {
printf("Insert title: ");
scanf("%s", titolo->nome);
LISTinsert(list, titolo);
}
return list;
}
in caricaFile you save all the read titles in the same location (titolo->nome) so only the last is finaly memorized because you save all the time the same pointer (titolo) in list.
You need to duplicate the val when you save it in the list, rather than to just do x->val = val; in NEW, for instance :
link NEW(TITOLO val, link next) {
link x = malloc(sizeof *x);
x->next = next;
TITOLO copy = malloc(sizeof (struct tit));
strcpy(copy->nome, val->nome);
x->val = copy;
return x;
}
Execution with the new definition of NEW :
Insert title: aze
Insert title: qsd
Insert title: wxc
Insert title: iop
aze iop qsd wxc
It is a very bad idea to make typedef being pointer like TITOLO is because this put the reader in error thinking it is a value rather than a pointer
Related
For context, I'm a new programmer to C and I wanted to make a toy implementation of a dictionary/map from a 'Person' struct to an integer. I'm using separate chaining, so I have a hash table of linked list pointers.
So far, I've been able to add one value to the linked list just fine, but when I call the function to get the value for the Person key I'm using, the memory at one of my nodes seems to get overwritten.
More info if it's helpful, using a singly linked list with one sentinel node at the head and a tail reference.
New to StackOverflow, so I can't actually embed the image, but pictured on left is the HashTable at the beginning of the function call, when nothing has been changed. The relevant stuff is the expanded part of the Variables menu, which shows that at position 58 is a pointer to 0x61f8e0, the linked list. The linked list has a head pointer to 0x61f760, which is the sentinel value, and a tail pointer to 0x61f864, currently pointing to a Node with the value (3) for a Person named Robert who's 36 years old. The tail pointer's next field points to 0x0 (not pictured), like intended. The picture follows: https://i.stack.imgur.com/F9EJ9.png
This is what happens as soon as the first statement (which hashes the Person pointer very naively) is executed: https://i.stack.imgur.com/UJvGy.png. As you'll see, the value is now some random long number, the intrinsic age is now 1 instead of 36, the saved name is now gibberish, and worst of all the next pointer now points somewhere completely random (0x61fb10).
The function in question follows.
int tableGet(HashTable t, Person key) {
int position = hash(&key) % 100;
List* listLoc = t.table[position];
if ((int) listLoc == 0) {
return -1;
}
Node curr = *(listLoc -> head);
while (curr.next != NULL) {
if (curr.savedAge == key.age && curr.savedName == key.name) {
return curr.val;
}
curr = *curr.next;
}
return -1;
}
Here is the hash function, in case that's what's causing the problems.
int hash(Person* p) {
int sum;
Person person = *p;
int i = 0;
char nameChar = person.name[i];
while (nameChar != '\0'){
sum += (int) nameChar;
i += 1;
nameChar = person.name[i];
}
return (int) (person.age + sum);
}
And just because why not, here's all of the short amount of code I've written for this.
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdbool.h>
typedef struct Node {
int val;
int savedAge;
char savedName[100];
struct Node* next;
} Node;
typedef struct {
int age;
char name[100];
} Person;
typedef struct {
Node* head;
Node* tail;
} List;
typedef struct {
List* table[100];
} HashTable;
int hash(Person*);
Person person(int age, char name[]) {
Person p;
p.age = age;
strcpy(p.name, name);
return p;
}
Node node(int val, Person p, Node* next) {
Node n;
n.val = val;
strcpy(n.savedName, p.name);
n.savedAge = p.age;
n.next = next;
return n;
}
List list() {
List l;
Node head = node(-1, person(0, "SENTINEL"), NULL);
l.head = &head;
l.tail = l.head;
return l;
}
void listAdd(List* l, Node n) {
Node* newTailPtr = &n;
l -> tail -> next = newTailPtr;
l -> tail = newTailPtr;
}
HashTable table() {
int table[100] = {0};
HashTable t;
memcpy(t.table, table, sizeof table);
return t;
}
HashTable tableAdd(HashTable t, Person key, int val) {
int num = hash(&key) % 100;
List* loc = t.table[num];
if ((int) loc == 0) {
List newList = list();
t.table[num] = &newList;
}
listAdd((List*) t.table[num], node(val, key, NULL));
return t;
}
int tableGet(HashTable t, Person key) {
int position = hash(&key) % 100;
List* listLoc = t.table[position];
if ((int) listLoc == 0) {
return -1;
}
Node curr = *(listLoc -> head);
while (curr.next != NULL) {
if (curr.savedAge == key.age && curr.savedName == key.name) {
return curr.val;
}
curr = *curr.next;
}
return -1;
}
int hash(Person* p) {
int sum;
Person person = *p;
int i = 0;
char nameChar = person.name[i];
while (nameChar != '\0'){
sum += (int) nameChar;
i += 1;
nameChar = person.name[i];
}
return (int) (person.age + sum);
}
int main() {
Person bob = person(36, "Robert");
printf(bob.name);
printf("\n");
HashTable tab = table();
tab = tableAdd(tab, bob, 3);
printf("Added Robert to table as 3\n");
int val = tableGet(tab, bob);
if (val == 3) {
printf("Success!\n");
} else {
printf("Failure, val is %d\n", val);
}
return 0;
}
I have create two struct and two variables n1 and n2, here is the result when runnig code below:
10-nick-90.500000
20-wilson-100.500000.
What I want:
1.take elemtype variable values and put them into struct LNODE.
2.print values by using printlist(LNDOE* head).
Please help me.
#include <stdio.h>
#include <stdlib.h>
typedef struct elemtype{
int no;
char* name;
double score;
}elemtype;
typedef struct tagnode{
elemtype data;
struct tagnode* next;
}LNODE;
elemtype* print_elemtype(elemtype* data_to_print){
elemtype* p = data_to_print;
if ( p != NULL){
printf("%d-%s-%lf", p->no,p->name,p->score);
}
printf("\n");
return NULL;
}
void creat_link(LNODE** head, int n, void (*input)(elemtype*)){
LNODE* s;
*head = (LNODE*)malloc(sizeof(LNODE));
(*head)->next = NULL;
for(;n>0;n--){
s = (LNODE*)malloc(sizeof(LNODE));
input(&(s->data));
s->next = (*head)->next;
(*head)->next = s;
}
}
int main()
{
elemtype* n1 = malloc(sizeof(elemtype));
elemtype* n2 = malloc(sizeof(elemtype));
n1->name = "nick";
n1->no = 10;
n1->score = 90.5;
n2->name = "wilson";
n2->no = 20;
n2->score = 100.5;
elemtype** s1,**s2;
s1 = &n1;
s2 = &n2;
print_elemtype(n1);
print_elemtype(n2);
free(n1);
free(n2);
n1 = NULL;
n2 = NULL;
}
Crawl in your linked list with kind of code like this :
int main(void)
{
struct_type *list = allocat_struct_type();
struct_type *list_copy = list; // always copy the head of the list when your manipulating the list
while (list_copy != NULL) {
printf("%s", list_copy->text_to_display); // text_to_display must be an str type in this exemple
list_copy = list_copy->next;
}
}
if your still having problem precise where and when.
Feel free to ask me again i hope it help you i tried to understand what you would know but it's very blurry.
This program asks for two data ( a name and an id), and then stores them in a linked list and then it should output all the data one by one.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct Test
{
char name[16];
int id;
};
typedef struct Node {
struct Test structure;
struct Node * next;
}TNode;
typedef TNodo* Node;
void NewDatainNode(struct Test p, Node *pp)
{
TNode *temp;
temp = malloc(sizeof(struct Node));
temp->structure = p;
temp->next = NULL;
*pp = temp;
}
void ReadData(struct Test * p)
{
printf("\nName:");
scanf(" %s", p->name);
printf("\nID:");
scanf(" %d", &p->id);
}
void ViewNodes(Node node)
{
while(node != NULL)
{
printf("%s %d\n",node->structure.name, node->structure.id);
node = node->next;
}
}
int Menu()
{
int c;
printf("\n\tM E N U ***\n"
"1 - New\n"
"2 - Print\n"
"0 - Exit\n"
"\n>> ");
scanf(" %d", &c);
return c;
}
int main()
{
int c;
struct Prova test;
struct Node * list = NULL;
do {
c = Menu();
switch (c)
{
case 1: ReadData(&test);
NewDatainNode(test, &list);break;
case 2: ViewNodes(lista); break;
default: c = 0;
}
} while (c != 0);
return 0;
}
The problem is that it outputs only the last data that has been inserted, what should i do to make it output all the data retrieved in function NuovaPrenotazione()?.
UPDATE: Now the function NewDatainNodes() looks like this:
void NewDatainNode(struct Test p, Node *pp)
{
TNode *temp;
temp = malloc(sizeof(struct Node));
temp->structure = p;
temp->next = *pp;
*pp = temp;
}
Thanks to Sandeep for the correction, the function works even without a return value,the problem was temp->structure = NULL.
This question is exactly duplicate copy of a question posted yesterday and marked as duplicate. But, here is your working code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
struct Prova
{
char nome[16];
int id;
};
typedef struct Node {
struct Prova struttura;
struct Node * next;
}TNodo;
typedef TNodo* Nodo;
Nodo NuovaPrenotazione(struct Prova p, Nodo pp)
{
Nodo temp;
temp = (Nodo)malloc(sizeof(struct Node));
temp->struttura = p;
temp->next = pp;
pp = temp;
return pp;
}
void LeggiPrenotazione(struct Prova * p)
{
printf("\nNome Cliente:");
scanf(" %s", p->nome);
printf("\nID:");
scanf(" %d", &p->id);
}
void Visualizza(struct Prova p)
{
printf("%s %d\n", p.nome, p.id);
}
void VisualizzaPrenotazione(Nodo nodo)
{
while(nodo != NULL)
{
Visualizza(nodo->struttura);
nodo = nodo->next;
}
}
int Menu()
{
int scelta;
printf("*** M E N U ***\n"
"1 - New\n"
"2 - Print\n"
"0 - Esci\n"
"Scelta --> ");
scanf(" %d", &scelta);
return scelta;
}
int main()
{
int scelta;
struct Prova test;
Nodo lista = NULL;
do {
scelta = Menu();
switch (scelta)
{
case 1: LeggiPrenotazione(&test);
lista = NuovaPrenotazione(test, lista);
break;
case 2: VisualizzaPrenotazione(lista); break;
default: scelta = 0;
}
} while (scelta != 0);
return 0;
}
Update: Here is the output
I'm trying to recurvisely implement an insert function in C, but I'm having all sorts of issues. I'm in an intro CS class and they started throwing C labs at us before we had ever worked with it before. Part of the issue is that the list pointer is not being recognized as NULL and I'm also pretty sure that I'm using malloc improperly.
#include <stdio.h>
#include <stdlib.h>
#define True 1
#define False 0
typedef int BOOLEAN;
struct Node{
int value;
struct Node *next;
};
void insert(int x, struct Node **pL){
printf("insert\n");
if(*pL == NULL){
printf("inside if\n");
struct Node *pN;
pN = (struct Node*) malloc(sizeof(struct Node));
(*pN).value = x;
(*pN).next = NULL;
return;
}
if (*pL != NULL){
printf("inside else\n");
insert(x, &(((*pL)->next)));
}
printf("end insert\n");
};
void printList(struct Node *L){
while (L != NULL){
printf("%d", (*L).value);
printList((*L).next);
}
return;
};
main(){
printf("main\n");
struct Node* L;
//L).next = NULL;
int i;
printf("for loop\n");
for (i = 3; i < 20; i+=2){
printf("%d\n", i);
insert(i, &L);
}
printList(L);
};
First of all, in main you need to initialise L:
struct Node* L = NULL;
Second, in insert when you allocate the new node pN, you are not assigning it to the pL, i.e., it doesn't get inserted. Put this right before the return; in insert:
*pL = pN;
(Also you could remove the return and change if (*pL != NULL) into else.)
Then, in printList you are both iterating with a while loop, and with recursion. Choose one, not both, e.g.,:
while (L) {
printf("%d\n", L->value);
L = L->next;
}
Also, throughout your code you can replace (*pointer_to_struct).field with pointer_to_struct->field for better style.
Write a function that rearranges a linked list to put the nodes in even positions after the nodes in odd positions in the list, preserving the relative order of both the evens and the odds.
I found this problem in the book Algorithm in c writtern by Sedgewick. I have tried but failed. I trid to put all nodes in even positions on another linked list. It's grateful for you to help me. A good idea is enough. Thanks :).
This is my Code in C.
/*
* File: rearranges.c <Exercise 3.36>
* Note: Write a function that rearranges a linked list to put the nodes in even
* positions after the nodes in odd positions in the list, preserving the
* relative order of both the evens and the odds.
* NOTICE: I think it's necessary to use linked list with a dummy head.
* Time: 2013-10-26 10:58
*/
#include <stdio.h>
#include <stdlib.h>
#define LEN 11
typedef struct node *link;
struct node {
int item;
link next;
};
/* Traverse a linked list with a dummy head. */
void traverse(link t) {
link x = t->next;
while (x != NULL) {
printf("%d ", x->item);
x = x->next;
}
putchar('\n');
}
/* Detach even positon nodes from a linked list. */
link detach(link t) {
link u = malloc(sizeof(*u));
link x = t, y = u;
/* x is odd position node. We should ensure that there's still one even
* position node after x. */
while (x != NULL && x->next != NULL) {
y->next = x->next;
x->next = x->next->next;
x = x->next;
y = y->next;
y->next = NULL;
}
return u;
}
/* Combine two linked list */
link combine(link u, link t) {
link x = u;
link y = t->next;
while (y != NULL) {
link n = y->next;
y->next = x->next;
x->next = y;
x = x->next->next;
y = n;
}
return u;
}
/* The function exchanges the position of the nodes in the list. */
link rearranges(link t) {
link u = detach(t);
link v = combine(u, t);
return v;
}
int main(int argc, char *argv[]) {
int i;
link t = malloc(sizeof(*t));
link x = t;
for (i = 0; i < LEN; i++) {
x->next = malloc(sizeof(*x));
x = x->next;
x->item = i;
x->next = NULL;
}
traverse(t);
traverse(rearranges(t));
return 0;
}
curr=head;
end=lastOfList;//last node if size of list is odd or last-1 node
for(int i=1;i<=listSize()/2;i++)
{
end->next=curr->next;
end=end->next;
end->next=null;
if(curr->next!=null)
if((curr->next)->next!=null)
curr->next=(curr->next)->next;
curr=curr->next;
}
You can implement a recursive solution where each call returns an updated node that will serve as the new next reference for the upper caller. We just have to go down the list until we find the last element, and then move every even node to the end of the list, and update the reference to the last element. Here's my solution (please try to do it yourself before looking at my and other solutions):
struct node {
int val;
struct node *next;
};
struct node *reorder_aux(struct node *l, int count, struct node **last);
struct node *reorder(struct node *l) {
struct node *x;
if (l == NULL)
return NULL;
return reorder_aux(l, 1, &x);
}
struct node *reorder_aux(struct node *l, int count, struct node **last) {
struct node *n;
if (l->next == NULL) {
*last = l;
return l;
}
n = reorder_aux(l->next, count+1, last);
if (count & 1) {
l->next = n;
return l;
}
else {
(*last)->next = l;
l->next = NULL;
*last = l;
return n;
}
}
At each step, if the current node l is an even node (as determined by count), then we append this node to the end, and tell the upper caller that its next pointer shall be updated to our next (because our next will be an odd node). In case we're an odd node, we just have to update our next pointer to whatever the recursive call returned (which will be a pointer to an odd node), and return the current node, since we will not move ourselves to the end of the list.
It's a nice exercise!
#include <stdio.h>
struct list {
struct list *next;
int ch;
};
void swap_odd_even (struct list **pp)
{
struct list *one, *two ;
for( ; (one = *pp) ; pp = &one->next) {
two = one->next;
if (!two) break;
*pp = two;
one->next = two->next;
two->next = one;
}
}
struct list arr[] =
{ {arr+1, 'A'} , {arr+2, 'B'} , {arr+3, 'C'} , {arr+4, 'D'}
, {arr+5, 'E'} , {arr+6, 'F'} , {arr+7, 'G'} , {arr+8, 'H'}
, {arr+9, 'I'} , {arr+10, 'J'} , {arr+11, 'K'} , {arr+12, 'L'}
, {arr+13, 'M'} , {arr+14, 'N'}, {arr+15, 'O'} , {arr+16, 'P'}
, {arr+17, 'Q'} , {arr+18, 'R'} , {arr+19, 'S'} , {arr+20, 'T'}
, {arr+21, 'U'} , {arr+22, 'V'}, {arr+23, 'W'} , {arr+24, 'X'}
, {arr+25, 'Y'} , {NULL, 'Z'} };
int main (void) {
struct list *root , *ptr;
root = arr;
for (ptr=root ; ptr; ptr = ptr->next ) {
printf( "-> %c" , ptr->ch );
}
printf( "\n" );
printf( "Swap\n" );
swap_odd_even ( &root);
for (ptr=root ; ptr; ptr = ptr->next ) {
printf( "-> %c" , ptr->ch );
}
printf( "\n" );
return 0;
}
In the following, every time swap_nodes is called another odd sinks to the last sunk odd. The evens are grouped together on each iteration and they bubble up to the end of the list. Here is an example:
/*
[0]-1-2-3-4-5
1-[0-2]-3-4-5
1-3-[0-2-4]-5
1-3-5-[0-2-4]
*/
#include <stdio.h>
#include <stdlib.h>
#define LIST_LENGTH 10
struct node{
int id;
struct node *next;
};
void print_list(struct node *current)
{
while(NULL != current){
printf("node id = %d\n",current->id);
current = current->next;
}
printf("Done\n");
}
struct node *swap_nodes(struct node *head_even, struct node *tail_even, struct node *next_odd)
{
tail_even->next = next_odd->next;
next_odd->next = head_even;
return next_odd;
}
struct node *reorder_list(struct node *head)
{
struct node *head_even;
struct node *tail_even;
struct node *next_odd;
struct node *last_odd;
if(NULL == head->next){
return head;
}
head_even = head;
tail_even = head;
next_odd = head->next;
last_odd = head->next;
head = swap_nodes(head_even, tail_even, next_odd);
if(NULL != tail_even->next){
tail_even = tail_even->next;
}
while (NULL != tail_even->next) {
next_odd = tail_even->next;
last_odd->next = swap_nodes(head_even, tail_even, next_odd);
last_odd = last_odd->next;
if(NULL != tail_even->next){
tail_even = tail_even->next;
}
}
return head;
}
int main(void)
{
int i;
struct node *head = (struct node *) malloc(LIST_LENGTH*sizeof(struct node));
struct node *mem = head;
if(NULL == head){
return -1;
}
struct node *current = head;
for(i=0;i<LIST_LENGTH-1;i++){
current->next = current + 1;
current->id = i;
current = current->next;
}
current->next = NULL;
current->id = i;
head = reorder_list(head);
print_list(head);
free(mem);
return 0;
}