Not working single linked list 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;
}

[casts are bad for malloc - see this](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc) — Ed Heal, Dec 11 '16 at 14:54
The use of `head` as a function parameter hides it from the other function. You should have played with double pointers some more. — Jongware, Dec 11 '16 at 14:56
Possible duplicate of [Passing by reference in C](http://stackoverflow.com/questions/2229498/passing-by-reference-in-c) — Siguza, Dec 11 '16 at 14:57
Inside `add` you change the value of `head`. Try to find a way how to pass the new value of `head` up the caller. — alk, Dec 11 '16 at 15:01
What happens to `temp` is `head` is not null - kinda disappears - memory leak — Ed Heal, Dec 11 '16 at 15:01
*head=temp;* in the add function doesn't do what you think is doing. The *head* pointer you are passing in the add function is actually a copy of you header. So *head=temp;* doesn't change the original but the copy one. You need to send the address of your header — γηράσκω δ' αεί πολλά διδασκόμε, Dec 11 '16 at 15:01
`head=temp;` do not do what you intended, it do not actually swap pointer for main head — Logman, Dec 11 '16 at 15:01
I know all these things. I just can't make double pointers work. Sad, but true. — , Dec 11 '16 at 15:11
Draw a diagram - I find this helps. Pointers being an arrow and the variable being a box — Ed Heal, Dec 11 '16 at 15:14

UrbiJr · Answer 1 · 2016-12-11T15:32:07.040

0

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.

edited Dec 11 '16 at 15:32

answered Dec 11 '16 at 15:19

UrbiJr

133
1
2
20

I tried such way. I get segfault. I am very bad today. I am so confused. – Dec 11 '16 at 15:21
Then please post your code and the errors you get, so that it will be simplier for us to help you. – UrbiJr Dec 11 '16 at 15:23
In this actual code I still see that you use `head` instead of `*head` in your `add` function. – UrbiJr Dec 11 '16 at 15:25
`head = (struct node **) head;` also remove this. – UrbiJr Dec 11 '16 at 15:27
Could you add the output with this final version of your code? – UrbiJr Dec 11 '16 at 15:52

score 0 · Answer 2 · answered Dec 11 '16 at 15:31

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!

well, it is great idea. But I have played with double pointers too much. So, I want make them work too. and what is bad with logic in else? And I know that my code should not work! I just gave an example for u to play :) — , Dec 11 '16 at 15:41

score 0 · Accepted Answer · edited Dec 11 '16 at 16:17

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.

Not working single linked list C

3 Answers3