Reputation: 11
#include <stdio.h>
#include <stdlib.h>
#define UINT unsigned int
struct item
{
UINT time ; // time in tics which is used to order the list.
UINT id ; // penguin id number.
UINT event ; // event this item describes.
struct item *next ; // pointer to next item in list.
};
struct item *head=NULL;
struct item *trail=NULL;
void link_list(UINT time, UINT id, UINT event)
{
struct item *t=NULL;
static int i=0;
if(i==0)
{
head->time=time;
head->id=id;
head->event=event;
trail=head;
trail->next=NULL;
i++;
printf("Hello Word\n");
}
t=malloc(sizeof(struct item));
trail->next=t;
trail->time=time;
trail->id=id;
trail->event=event;
trail=t;
trail->next=NULL;
if(i!=0)
printf("I am \n");
}
int main(int argc, char *argv[])
{
UINT x[3]={4,5,7}, y[3]={40,50,60}, z[3]={100,900,500};
int i=0;
head=malloc(sizeof(struct item));
trail=head;
link_list(x[0], y[0], z[0]);
link_list(x[1], y[1], z[1]);
link_list(x[2], y[2], z[2]);
struct item *f=NULL;
f=head;
do
{
printf("trail: %d %d %d\n", f->time, f->id, f->event);
f=f->next;
}
while(f!=NULL);
return 0;
}
Good day,
I'm currently having a logical problem with my code implementation of my linked list logically. This code is the framework I am using to integrate it into a larger program which will use linked list so I need to get this right.
Basically what is happening is when I finally reach the do while loop which I use as a debugging line to see the contents of a linked list I will get this output on my command line:
trail: 4 40 100
trail: 5 50 900
trail: 7 60 500
trail: 0 0 0
I am expecting the output to be like this:
trail: 4 40 100
trail: 5 50 900
trail: 7 60 500
I have excluded the other printf's in my code as they are just used to check if I am indeed going through my functions properly. Also this might be unrelated but is there a better debugger for c under linux? Because the built in debugger goes crazy when it steps into a malloc command so I have to debug all programs inside my head. :(
Upvotes: 0
Views: 169
Reputation: 66204
You're appending a new node to the tail, then writing in your datum to the tail node, then moving the tail pointer to the new node:
t=malloc(sizeof(struct item));
trail->next=t;
trail->time=time;
trail->id=id;
trail->event=event;
trail=t;
trail->next=NULL;
t is set to a new node. the next pointer of the current list tail is set to point to t the time, id, and event fields of the node pointed by tail are written. tail reference is now moved to t, which still contains no data.
There are a plethora of things that can improve this code, starting with using modular design for a "list" management, but this is your immediate problem.
Upvotes: 1
Reputation: 619
If by built-in debugger you mean gdb, you can tell it to step over malloc(); if your program crashes in malloc, though, you have run into a memory allocation or use bug in your program.
One apparent bug is that you do not initialize "t" correctly in link_list(), and in particular t->next is garbage, and your program will probably crash when it tries to follow it. I think you mean to set t->time, t->id etc., not trail->time.
The best way to append to a linked list is usually to have a separate function that returns a new list item with all the fields initialized, and then in the routine to append, just manipulate pointers. It also helps if your list head is just a pointer, not a struct, but that's a style issue.
Best of all is to use a library that supplies linked lists for you, but that's no good if you are learning about linked lists!
Upvotes: 1