How to access data from a Linked List?

This is a discussion on How to access data from a Linked List? within the C Programming forums, part of the General Programming Boards category; Hi! I believe it must be quite simple to do, and I'm not finding it on the forum. I'd like ...

  1. #1
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    Sep 2010
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    8

    Question How to access data from a Linked List?

    Hi!

    I believe it must be quite simple to do, and I'm not finding it on the forum. I'd like to access a specific variable in my code which is in a Linked List.

    I've tried to use something like struct_name[position].variable_name, but It's not working. On the code below, I tried to print its value...

    What am I supposed to do?

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    
    struct list {
        int info;
        struct list* next;
    };
       
    typedef struct list List;
    
    List* initialize (void)
    {
         return NULL;
    }
        
    List* insert (List* l, int i)
    {
           List* new = (List*) malloc(sizeof(List));
           new->info = i;
           new->next = l;
           return new;
     }
        
    void print (List* l)
     {
           List* p;   
           for (p = l; p != NULL; p = p->next)
    
                printf("info = %d\n", p->info);
     }
        
    int main(int argc, char *argv[])
    {
        
        l = initialize();  
        l = insert(l, 23); 
        l = insert(l, 45);
      
         print(l);
         printf("%d\n",l[0].info); //<------------------------ Here
      
    
         system("PAUSE");    
         return 0;
    Thanks =)

  2. #2
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    i've tried to use something like struct_name[position].variable_name, but It's not working. On the code below, I tried to print its value...
    structures are NOT array, so struct-name[position], is wrong.

    besides it is a LINKED list where 1 structure points to another one SINGLE structure, so they are not even "similar" to arrays.

    Now, if you look carefully you are dealing here with POINTERS TO structure.

    now if p is a pointer to a structure, you access its members by (*p).member_name.
    or more simply by p->member_name, which is just a shorthand for the previous version.

    so here it should be l->info.
    now, l is a pointer to the FIRST structure, so you have to enter the loop by assigning to l the pointer to the next structure.
    .
    In order to terminate the loop, you make sure that it stops when the structure pointer l is equal to null, i.e. 0, yes address zero, which is a sign that we reached one structure beyond the "meaningful" structures, and in fact this is exactly what you defined above for function print().
    Last edited by alecodd; 01-25-2011 at 08:23 PM.

  3. #3
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    Sep 2010
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    8
    uhmm

    So the only way is to use a loop to access one by one...
    Thanks for your explanation, alecodd!

    This code is not mine, actually. That's why one print routine works and the another one doesn't. I'm using it to understand how all this stuff work, and than work on my project =)

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