I got a linked list, which should save the Outcome (W or L) and the gained/lost points for each match. All good so far, but I'm getting trouble when the head does not exist/is empty. I also realized I have a pretty bad overview of how to implement linked lists, anyone got good and understandable resources? Anyway this is my code:
#include <stdio.h>
#include <stdlib.h>
struct node {
int point;
char outcome;
struct node *next;
};
void add(struct node *data){
if(data == NULL){
data = malloc(sizeof(struct node));
printf("Outcome and points?\n");
int point;
char outcome;
scanf("%c %d",&outcome,&point);
fgetc(stdin);
data->point=point;
data->outcome=outcome;
data->next=NULL;
}else{
struct node *current= data;
while(current->next != NULL){
current = current->next;
}
current->next = malloc(sizeof(struct node));
current=current->next;
printf("Outcome and points?\n");
int point;
char outcome;
scanf("%c %d",&outcome,&point);
fgetc(stdin);
current->point=point;
current->outcome=outcome;
current->next=NULL;
}
}
void print(struct node *data){
struct node *current = data;
while(current != NULL){
printf("%c with %3d\n",current->outcome,current->point);
current = current->next;
}
}
int main()
{
struct node *head=NULL;
add(head);
add(head);
add(head);
print(head);
}
Any help would be appreciated :)
When you execute:
void add(struct node *data){
if(data == NULL){
data = malloc(sizeof(struct node));
the value of head does not change in the calling function.
Suggest a change of strategy.
struct node* add(struct node *head)
{
if(head == NULL){
head = malloc(sizeof(struct node));
printf("Outcome and points?\n");
int point;
char outcome;
scanf("%c %d",&outcome,&point);
fgetc(stdin);
head->point=point;
head->outcome=outcome;
head->next=NULL;
}else{
struct node *current= head;
while(current->next != NULL){
current = current->next;
}
current->next = malloc(sizeof(struct node));
current=current->next;
printf("Outcome and points?\n");
int point;
char outcome;
scanf("%c %d",&outcome,&point);
fgetc(stdin);
current->point=point;
current->outcome=outcome;
current->next=NULL;
}
return head;
}
And, then change the usage:
int main()
{
struct node *head = add(NULL);
add(head);
add(head);
print(head);
}
You can simplify the code by starting the list with an anchor node. An anchor node is a node that is only used for its next pointer. In the code below, the call to calloc creates the anchor node, and sets all of the fields in the anchor to 0. In other words, a node is created with next == NULL.
Note that when printing the list, the for loop starts by skipping the anchor node with for (list = list->next;...)
#include <stdio.h>
#include <stdlib.h>
struct node
{
int point;
char outcome;
struct node *next;
};
void add( struct node *list )
{
struct node *data;
data = malloc(sizeof(struct node));
data->next = NULL;
printf("Outcome and points?\n");
scanf("%c %d",&data->outcome,&data->point);
fgetc(stdin);
while (list->next != NULL)
list = list->next;
list->next = data;
}
void print( struct node *list )
{
for (list = list->next; list != NULL; list = list->next)
printf("%c with %3d\n", list->outcome, list->point);
}
int main()
{
struct node *head = calloc( 1, sizeof(struct node) );
add(head);
add(head);
add(head);
print(head);
}
Side note: I've omitted some error checking to keep things simple, you should really be checking the return values from calloc, malloc, and scanf and handle any errors. And, of course, you should free all of the nodes at the end.
Related
I'm new to programming in C and taking a course. I'm having trouble with one of the tasks I'm practicing. I'm supposed to Write a program that creates a linked list of 10 characters, then creates a copy of the list in reverse order. I have written (mostly copied) a code, but it only reverses the contents of my linked list, doesn't copy them to a new linked list in reverse order. It's also not working with letters even though I'm using char data type. works fine with numbers.
Here's my code:
#include <stdio.h>
#include <malloc.h>
struct Node
{
char data;
struct Node *next;
};
static void reverse(struct Node **head_ref)
{
struct Node *previous = NULL;
struct Node *current = *head_ref;
struct Node *next;
while (current != NULL)
{
next = current->next;
current->next = previous;
previous = current;
current = next;
}
*head_ref = previous;
}
void push(struct Node **head_ref, char new_data)
{
struct Node *new_node =
(struct Node *)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void printList(struct Node *head)
{
struct Node *temp = head;
while (temp != NULL)
{
printf("%d ", temp->data);
temp = temp->next;
}
}
int main()
{
struct Node *head = NULL;
char element = NULL;
printf("Enter 10 characters:\n");
for (int i = 0; i <= 9; i++)
{
scanf_s("%d", &element);
push(&head, element);
}
printf("Given linked list\n");
printList(head);
reverse(&head);
printf("\nReversed Linked list \n");
printList(head);
getchar();
}
This for loop
for (int i = 0; i <= 9; i++)
{
scanf_s("%d", &element);
push(&head, element);
}
invokes undefined behavior because there is used an incorrect conversion specifier %d with an object of the type char,
You need to write
for (int i = 0; i <= 9; i++)
{
scanf_s( " %c", &element, 1 );
push(&head, element);
}
Pay attention to the blank before the conversion specifier %c in the format string. This allows to skip white space characters in the input stream.
As for the function then it can be declared and defined the following simple way using the function push that you already defined
struct Node * reverse_copy( const struct Node *head )
{
struct Node *new_head = NULL;
for ( ; head != NULL; head = head->next )
{
push( &new_head, head->data );
}
return new_head;
}
And in main you can write something like
struct Node *second_head = reverse_copy( head );
Take into account that the function push would be more safer if it would process the situation when memory allocation for a node failed.
To create a copy in reverse order, create a new list with the same values as the original list but prepend the new nodes using the push function.
Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
struct Node {
char data;
struct Node *next;
};
void prepend(struct Node **head_ref, char new_data) {
struct Node *new_node = (struct Node *)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
void append(struct Node **head_ref, char new_data) {
struct Node *new_node = (struct Node *)malloc(sizeof(struct Node));
struct Node *node = *head_ref;
new_node->data = new_data;
new_node->next = NULL;
if (!node) {
*head_ref = new_node;
} else {
while (node->next)
node = node->next;
node->next = new_node;
}
}
void printList(const struct Node *head) {
const struct Node *temp = head;
while (temp != NULL) {
printf("%c ", temp->data);
temp = temp->next;
}
printf("\n");
}
struct Node *copy_reverse(struct Node *list) {
struct Node *new_list = NULL;
while (list) {
prepend(&new_list, list->data);
list = list->next;
}
return new_list;
}
void freeList(struct Node *list) {
while (list) {
struct Node *node = list;
list = list->next;
free(node);
}
}
int main() {
struct Node *head = NULL;
char element;
printf("Enter 10 characters:\n");
for (int i = 0; i < 10; i++) {
scanf_s("%c", &element);
push(&head, element);
}
printf("Given linked list\n");
printList(head);
struct Node *copy = copy_reverse(head);
printf("\nReversed Linked list \n");
printList(copy);
freeList(head);
freeList(copy);
getchar();
}
You're almost there. All it needs is one tweak. In reverse, you need to create a new copy of the current node and use that instead. Also, since you'll be ending up with a second list and not altering the original, you should return the new list from reverse.
static struct Node* reverse(const struct Node* head_ref)
{
struct Node* previous = NULL;
const struct Node* current = head_ref;
struct Node* copy;
while (current != NULL) {
copy = malloc(sizeof(*copy));
if (copy == NULL) {
// handle error
}
copy->data = current->data;
copy->next = previous;
previous = copy;
current = current->next;
}
return previous;
}
You can also make the loop prettier by converting it to a for loop.
for (current = head_ref; current != NULL; current = current->next) {
Finally, when you print out the list, you're using %d in the printf format string. %d will print the char as an integer. To print out the actual character, use %c instead.
This is my program in C which always inserts into a linked list at the end. But when I try to print the list elements, nothing is displayed. Here is the code :
#include<stdio.h>
#include<stdlib.h>
struct Node
{
int data;
struct Node *next;
};
void insert(struct Node *, int);
int main(void)
{
struct Node *head = NULL, *current;
int n, i, x, data;
scanf("%d", &n);
for(i = 0; i < n; i++)
{
scanf("%d", &data);
insert(head, data);
}
current = head;
while(current != NULL)
{
printf("%d ", current->data);
current = current->next;
}
}
void insert(struct Node *head, int data)
{
struct Node *newnode, *current = head;
newnode = (struct Node *)malloc(sizeof(struct Node));
newnode->data = data;
newnode->next = NULL;
if(head == NULL)
{
head = newnode;
}
else
{
while(current->next != NULL)
{
current = current->next;
}
current->next = newnode;
}
}
I cannot understand what might be the issue. Please help.
Your insert cannot modify head. Change it to
void insert(struct Node **head, int data)
and change it by
*head = newnode;
and call it like this
insert(&head, data);
Here, while you are passing the head pointer to your insert() function, it is not being updated in your main() function.
So, either declare your head pointer as global or return your head pointer and update it in your main() function.
In the below code I had taken the head pointer as global and removed the head pointer as your parameter from the insert() function.
Here is the code :-
#include<stdio.h>
#include<stdlib.h>
struct Node
{
int data;
struct Node *next;
};
struct Node *head=NULL;
void insert(int);
int main(void)
{
struct Node *current;
int n, i, x, data;
clrscr();
scanf("%d", &n);
for(i = 0; i < n; i++)
{
scanf("%d", &data);
insert(data);
}
current = head;
while(current != NULL)
{
printf("%d \n", current->data);
current = current->next;
}
getch();
return 0;
}
void insert(int data)
{
struct Node *newnode, *current = head;
newnode = (struct Node *)malloc(sizeof(struct Node));
newnode->data = data;
newnode->next = NULL;
if(head == NULL)
{
head = newnode;
}
else
{
while(current->next != NULL)
{
current = current->next;
}
current->next = newnode;
}
}
You need to pass the reference of the head pointer, then only the changes made to it will be visible.
You must declare your function like
void insert(struct Node **, int);
and also call it like
insert(&head, data);
also, make changes to function definition
void insert(struct Node **head, int data)
{
struct Node *newnode, *current = *head;
newnode = (struct Node *)malloc(sizeof(struct Node));
newnode->data = data;
newnode->next = NULL;
if(*head == NULL)
{
*head = newnode;
}
else
{
while(current->next != NULL)
{
current = current->next;
}
current->next = newnode;
}
}
You need to pass the head by reference as you are making changes to it that should be visible.
insert(head, data);
should become
insert(&head, data);
Also the function signature will change.
void insert(struct Node *head, int data)
should become
void insert(struct Node **head, int data)
Also make appropriate changes in the function.
Like,
current = *head;
Because you are passing the pointer by value. The function operates on a copy of the pointer, and never modifies the original.
Either pass a pointer to the pointer (i.e. a struct head **), or instead have the function return the pointer.
You can try running the following code which will give the output as null
printf("%s",head);
while(current != NULL)
{
printf("%d", current->data);
current = current->next;
}
I am weak in linked list so please help me:
My LinkedList is not displaying on screen(display function isn't working correctly) even the linked list is creating successfully and if i use node *head as a global pointer to structure and if i replaced list->head with head (which is the address of node *head) and LinkedList *list (in argument of function) with node *head, then it is printing the whole list but if i use list->head or list in place of head (which is the address of node *head) and declared LinkedList *list in term of node *head then it input the value in linked list but don't display the linked list.
The code which causing problem is following:
#include<stdio.h>
#include<malloc.h>
typedef struct node{
int data;
struct node *next;
}node;
typedef struct LinkedList_
{
node *head;
} LinkedList;
void create( LinkedList *list ){
char choice='y';
do{
node *newnode,*temp;
newnode=(node*)malloc(sizeof(node*));
printf("\nenter the data: ");
scanf("%d",&newnode->data);
newnode->next=NULL;
if(list==NULL){
list->head=newnode;
temp=newnode;
}
else{
temp=list->head;
while(temp->next!=NULL){
temp=temp->next;
}
temp->next=newnode;
}
printf("\ndo you want to continue(y or no)? ");
choice=getche();
}while(choice=='y');
}
void display( LinkedList *list ){
printf("\n[");
while(list->head!=NULL){
printf("%d,",list->head->data);
list->head=list->head->next;
}
printf("]");
}
void main(){
LinkedList *head=NULL;
create(head);
display(head);
}
fix like this
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
typedef struct node{
int data;
struct node *next;
}node;
typedef struct LinkedList_ {
node *head;
} LinkedList;
void create( LinkedList *list ){
char choice='y';
if(!list || list->head){//guard
fprintf(stderr, "Invalid call %s.\n", __func__);
return;
}
node *temp;
do{
node *newnode;
newnode = malloc(sizeof(*newnode));//or malloc(sizeof(node));
printf("\nenter the data: ");
scanf("%d", &newnode->data);
newnode->next = NULL;
if(list->head == NULL){
list->head = newnode;
temp = newnode;
} else {
temp = temp->next = newnode;//There is no need to follow the link
}
printf("\ndo you want to continue(y or no)? ");
choice=getche();
}while(choice=='y');
}
void display( LinkedList *list ){
node *curr = list->head;//head should not be changed
printf("\n[");
while(curr != NULL){
printf("%d,", curr->data);
curr = curr->next;
}
printf("]\n");
}
int main(void){
LinkedList head = { NULL };//requires entity
create(&head);
display(&head);
}
I am having some trouble adding integers to the end of my linked list. I am very new to C and had part of my program working properly (the push function). I want to return a pointer to a struct node, and I am not quite sure where I am going wrong in my append function.
~Thanks.
enter code here
//node.h
#ifndef NODE_H
#define NODE_H
struct node{
int val;
struct node *next;
};
int length(struct node *);
struct node* push(struct node *, int); //adds integer to front of list.
struct node* append(struct node *, int); //adds integer to back of list.
void print(struct node *, int);
#endif
//node.c
#include "./node.h"
#include<stdlib.h>
#include<stdio.h>
int length(struct node *current){
if(current->next != NULL)
return 1 + length(current->next);
else
return 1;
}
struct node* push(struct node *head, int num){
struct node *temp = malloc(sizeof(struct node));
temp->val = num;
temp->next = head;
head = temp;
temp = NULL;
return head;
}
struct node* append(struct node *current, int num){
if(current != NULL){
append(current->next, num);
}
else{
struct node* temp = malloc(sizeof(struct node));
temp->val = num;
temp->next = NULL;
current = temp;
return current;
}
}
void print(struct node* head, int size){
printf("The list is %i", size);
printf(" long \n");
struct node* temp;
temp = head;
while(temp != NULL){
printf("%d", temp->val);
printf(" ");
temp = temp->next;
}
printf(" \n");
}
//Main
#include "./node.h"
#include<stdlib.h>
#include<stdio.h>
int main(){
char ans[2];
int num;
struct node* head = NULL;
do{
printf("Enter a integer for linked list: ");
scanf("%d", &num);
head = append(head, num);
printf("Add another integer to linked list? (y or n) ");
scanf("%1s", ans);
}while(*ans == 'y');
print(head, length(head));
return 0;
}
I think what is missing is that the recursive part of the function needs to set current->next. This has the effect of setting every node's next pointer to what it was until you get to the end of the list, when it is set to the newly malloced node.
struct node* append(struct node *current, int num){
if(current != NULL){
current->next = append(current->next, num);
return current;
}
else {
struct node* temp = malloc(sizeof(struct node));
if (temp == NULL) abort();
temp->val = num;
temp->next = NULL;
return temp;
}
}
I have the following single linked list,I always get the length as 1,even if i push 3 elements,and always only one node is created.Please help.Thanks.
#include <stdio.h>
struct node
{
int data;
struct node *next;
};
void push(struct node **head,int data)
{
struct node *temp = (struct node*)malloc(sizeof(struct node));
temp->data=data;
if(*head == NULL)
{
*head=temp;
}
else
{
temp->next=*head;
*head=temp;
}
}
int length(struct node *head)
{
struct node *temp = head;
int count=0;
if(temp !=NULL)
{
count++;
printf("%d",temp->data);
temp=temp->next;
}
return count;
}
int main()
{
int a;
struct node *head=NULL;
push(&head,1);
push(&head,2);
push(&head,3);
a=length(head);
printf("%d",a);
return 0;
}
Replace if by while in the length function
In your length function, change this line:
if(temp !=NULL)
to this:
while(temp != NULL)
Have you noticed the structure of your length method? You are using an if statement where a loop would be appropriate. You are getting the answer of 1 because you are only executing the count ++ statement once.
Hope this helps.
The error comes from push() function. If head is not null you need to iterate through the list to the last node. And as said before while instead if
# include <stdlib.h>
void push(struct node **head,int data)
{
struct node *temp;
temp = malloc (sizeof *temp);
temp->data = data;
temp->next = *head;
*head = temp;
}