[Dev-C++] Single Linked List Problem-II

[<and...@su...>]

I forgot the "int i;" variable declaration, ok?
And this is a "C" program (aswering for the orientation "allways say if you
are working
with a C or C++ program").

-----Mensagem original-----
De: André Macário Barros [mailto:and...@su...]
Enviada em: quinta-feira, 5 de junho de 2003 22:58
Para: Dev-Cpp-Users
Assunto: Single Linked List Problem


Dear Users,

I need to discover a way to use the "free" command to correctly avoid 5
records dinamically allocated in a simple Single Linked List program.
Here are some pieces of it:

...

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

#define MAX 5

struct address
{
   char name[40];
   char phone[20];
   struct address *next;
};

void slstore(struct address *, struct address **, struct address **);

...
int main()
{

   struct address *info, *start, *last;

   init(&start);
   start=NULL;
   init(&last);
   last=NULL;
   for(i=0; i<MAX; i++)
   {
      init(&info);
      printf("Enter name: ");
      gets(info->name);
      printf("Enter phone: ");
      gets(info->phone);
      slstore(info, &start, &last);
   }
   ...
   system("PAUSE");
   return(0);
}

void slstore(struct address *i, struct address **start_arg,
             struct address **last_arg)
{
   if(*last_arg==NULL) *start_arg=i;
   else (*last_arg)->next=i;
   i->next=NULL;
   *last_arg=i;
}

void init(struct address **arg)
{
   *arg= malloc(1*sizeof(struct address));
   if(!(*arg))
   {
      printf("sem memória\n");
      system("PAUSE");
      exit(EXIT_FAILURE);
   }
}

***
The slstore function stores the new element in the last position. It´s
"next" record is filled
with the NULL address to mean that there aren´t more elements in the list.
At the end of the program 5 memory spaces of 64 bytes each one, are
allocated for each element.
How can I use the free command to return the memory for the computer?

Thanx
Andre