linked - list -> 2 var' in the struct ( add function)

Hi

I'm trying to build a link list with the enclosed code.
could you please help me to create an "add function" to my struct (list) with 2 components (int Y & int X)

Code:

typedef struct list
{
    int Y;
    int X;
    struct list *next;
}list;

void add ( list *head, int number)
{
    list *new_list;
    new_list=(list*)malloc(sizeof(list));
    new_list ?? number
}

thanks
Ilan

Well you could do this in the function body:

Code:

new_list = malloc (sizeof *new_list);
new_list->X = number;
new_list->Y = number;
new_list->next = NULL;
return new_list;

Among other things... I'm working on very little information. If that's wrong then I don't know what's right.

Originally Posted by ilans11il

Hi

I'm trying to build a link list with the enclosed code.
could you please help me to create an "add function" to my struct (list) with 2 components (int Y & int X)

Hello there,


If you use a void function, then you will need a pointer of pointer to your struct.

you can try the following:

Code:

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

typedef struct list
{
    int Y;
    int X;
    struct list *next;
}list;

typedef struct list * LinkedList;


void add (LinkedList* param_linkedListPtr, int param_number)
{
    if (*param_linkedListPtr == NULL)
    {
        *param_linkedListPtr = (LinkedList)malloc(sizeof(struct list));
        (*param_linkedListPtr)->X = param_number;
        (*param_linkedListPtr)->Y = param_number;
        (*param_linkedListPtr)->next = NULL;
    }
    else
    {
        LinkedList newList = (LinkedList)malloc(sizeof(list));
        newList->X = param_number;
        newList->Y = param_number;
        newList->next = NULL;
        
        // Add the new element at the end of the list
        // So let's go first, to the end of the list
        LinkedList tmpList = *param_linkedListPtr;
        while (tmpList->next != NULL)
            tmpList = tmpList->next;
        
        // Now add the new element at the end of the list
        LinkedList *tmpListPtr = &tmpList;
        (*tmpListPtr)->next = newList;
    }
}


void printList(LinkedList param_LinkedList)
{
    LinkedList tmpList = param_LinkedList;
    while (tmpList != NULL)
    {
        printf("%d \t %d \n", tmpList->X, tmpList->Y);
        tmpList = tmpList->next;
    }
}


int main(int argc, char *argv[])
{
    // start testing with an empty list
    LinkedList linkedList = NULL;
        
    add(&linkedList, 12);
    add(&linkedList, 14);
    add(&linkedList, 100);    
    add(&linkedList, 300);
    add(&linkedList, 400);
    add(&linkedList, 500);

    printList(linkedList);
    
    return 0;
}

Code:

$ gcc -Wall testscript.c -o testscript
$ ./testscript
12       12 
14       14 
100      100 
300      300 
400      400 
500      500 
$

Regards,
Dariyoosh

thanks ,but:
why to use :new_list = malloc (sizeof *new_list)
inteade of: new_list=(list*)malloc(sizeof(list));

thanks

There are two reasons:

1. You do not need to cast malloc in C since the void * type is assignable to any other pointer type.

See also, FAQ > Casting malloc - Cprogramming.com

2. The sizeof expression is not evaluated at run time, but at compile time, so you always will get the correct type from it. It's supposed to be a convenience and less prone to errors. Again, see the faq.