无法工作的单个链表C

Question

I am too bad now. My list is not working! I know that there is an issue of just coping my ptr into function, not actually using real one, but I can't understand how can I make this work as I want.

PS. I see also that if I make head as global value, it will be ok. But I want to get function, which I can call it specific list.

Here is function of adding element into a blamk list. I can't make even this function work. I tried to play with double pointers, but now I am here to ask some help.

   #include<stdio.h>
#include<stdlib.h>

struct node
{
    int data;
    struct node *next;
};

void add( int num,struct node * head )
{
    struct node *temp;
    temp=(struct node *)malloc(sizeof(struct node));
    temp->data=num;
    if (head== NULL)
    {
    head=temp;
    head->next=NULL;
    }
    else
    {
    temp->next=head;
    head=temp;
    }
}

int  main()
{
    struct node *head;
    head=NULL;
    add(20,head);
    if(head==NULL) printf("List is Empty\n");

    return 0;
}

UPD: MY own playing with double pointers:

    #include<stdio.h>
#include<stdlib.h>

struct node
{
    int data;
    struct node *next;
};

void add( int num, struct node **head )
{
    struct node **temp;
    temp=(struct node **)malloc(sizeof(struct node*));
    (*temp)->data=num;
    if (*head== NULL)
    {
    *head=*temp;
    (*head)->next=NULL;
    }
        else
{
(*temp)->next=head;
*head=temp;
}

}

int  main()
{
    struct node *head;
    head=NULL;
    add(20,&head);
    if(head==NULL) printf("List is Empty\n");

    return 0;
}

Answer 1

How do you expect it will work if you don't pass the address of head to the function?

Correct code:
add(20,&head)

And you also have to change your function signature with this:
void add(int num, struct node **head)

Also, to refer to your struct node pointer, again in the add function you'll want to change head with *head .

Please pay attention: **head points to *head , so everytime you want to apply changes to your pointer (not double pointer ) head , you have to tell it to the compiler by using *head .

Answer 2

How are you expecting it to work? It wont work unless you pass the pointer head itself, here you are just creating a copy of it. You can do this as well.

head = add(20,head);

Instead of just

add(20,head);

Add a

return head;

before ending the function

And don't forget to change the return type of your function like this

struct node* add( int num,struct node *head)

The updated code looks like this

#include <stdio.h>
#include <stdlib.h>

struct node
{
    int data;
    struct node *next;
};

struct node* add( int num,struct node *head)
{
    struct node *temp;
    temp=malloc(sizeof(struct node));
    temp->data=num;
    if (head==NULL)
    {
        head=temp;
        head->next=NULL;
    }
    else
    {
        // please change your logic
    }
    return head;
}

int  main()
{
    struct node *head;
    head=NULL;
    head = add(20,head);
    if(head==NULL) printf("List is Empty\n");

    return 0;
}

Cheers!

Answer 3

Change

struct node **temp;
temp=(struct node **)malloc(sizeof(struct node*));
(*temp)->data=num;
if (*head== NULL)
{
    *head=*temp;
    (*head)->next=NULL;
}

To

struct node *temp;
temp=(struct node *)malloc(sizeof(struct node));
(temp)->data=num;
if (*head== NULL)
{
    *head=temp;
    (*head)->next=NULL;
}

The reason you are getting a segmentation fault, is because in malloc, you allocated sizeof(struct node*) , which is basically enough size for a pointer and doesn't make any sense to do so.

Also, change the logic of the else in add, if you are planning to add in more nodes.