Trying to make linkedlist in C - c

Question

I am trying to make a struct in C that is a linked list. I am not really sure what is going wrong though. My errors are:
linked.c:6:2: error: unknown type name ‘linkedList’
linked.c: In function ‘makeList’:
linked.c:30:2: error: ‘first’ undeclared (first use in this function)
linked.c:30:2: note: each undeclared identifier is reported only once for each function it appears in
linked.c: In function ‘addToList’:
linked.c:36:9: error: used struct type value where scalar is required
linked.c:43:13: error: incompatible types when assigning to type ‘int *’ from type ‘linkedList’
if anybody can see what is wrong and explain it to me, it would be much appreciated. My code is below.
#include <stdio.h>
typedef struct linkedList
{
int first;
linkedList* rest;
} linkedList;
linkedList makeList(int a, int b, int c);
void addToList(linkedList* ll, int a);
int main()
{
linkedList ll = makeList(1,3,5);
addToList(&ll, 7);
addToList(&ll, 9);
return 0;
}
linkedList makeList(int a, int b, int c)
{
linkedList ll;
ll.first = a;
linkedList second;
second.first = b;
linkedList third;
third.first = c;
third.rest = NULL;
second.rest = &c;
first.rest = &b;
return first;
}
void addToList(linkedList* ll, int a)
{
while (*ll)
{
if (ll->rest == NULL)
{
linkedList newL;
newL.first = a;
newL.rest = NULL;
ll->rest = newL;
break;
} else
{
continue;
}
}
}

Answer 1

The C compiler doesn't have a complete typedef of linkedList before you attempt to use it in your struct. You have a couple of options:
typedef struct linkedList
{
int first;
struct linkedList* rest;
} linkedList;
Or:
typedef struct linkedList linkedList; // C allows this forward declaration
struct linkedList
{
int first;
linkedList* rest;
};
This is your starting point.
Additional problems include but are not limited to:
Your makeList function refers to variable first but it doesn't appear to be defined anywhere.
ll->rest = newL; assigned a type linkedList to a pointer to linkedList (linkedList *) you can't assign a value to a pointer-to-value. The compiler error message linked.c:43:13:... states this. It would need to be ll->rest = &newL;... HOWEVER...
newL is LOCAL to the function addToList, so you can't assign it's address to a persistent list item since it will go out of scope when the code leaves that block.
In addToList you are assigning pointer to integer to a variable that holds pointer to linkedList, e.g., second.rest = &c;.

Answer 2

here's a corrected version of your program :
#include <stdio.h>
#include <stdlib.h>
typedef struct linkedList
{
int first;
struct linkedList* rest; // add struct in the beginning
} linkedList;
linkedList* addToList(linkedList* ll, int a);
void go_trough(linkedList *ll); // here's an extra function to check
int main()
{
linkedList *ll ; // working with a pointer is easier and makelist is pointless work with add to list instead
ll = NULL; // initialize to NULL
ll = addToList(ll, 7);
ll = addToList(ll, 9);
go_trough(ll);
return 0;
}
linkedList* addToList(linkedList* ll, int a) // I didn't understand what you were trying to do so ... here's my version
{
if(!ll)
{
ll = malloc(sizeof(linkedList*)); //allocating enought space to hold the structure
ll->first = a;
ll->rest = NULL;
}
else
ll->rest = addToList(ll->rest , a);
return ll;
}
void go_trough(linkedList *ll)
{
if(ll)
{
printf("%d\n" , ll->first);
go_trough(ll->rest);
}
}

Answer 3

in makeList change
second.rest = &c;
first.rest = &b;
to
ll.rest = &second;
second.rest = &third;
in the original you were giving the adresses of the int variables instead of the linkedList nodes. also, you had a variable 'first' which was never declared, that's where one of errors were taking place.
also try declaring all your variables first, it makes it easier to read.

Answer 4

A few observations,
declare a struct name so that you can use it in the linkedList struct.
DRY - Don't Repeat Yourself, that is why the below ListNew() function is provided
use pointers, that is the whole point to building a linked list anyway,
your list uses one type of node, storing data and the list pointer,
name the pointer to the next node in the list whatever you want, how about 'next'?
name the thing that holds data anything you want, how about 'data'?
print the list, it will help figure out what is going on, :-)
a pointer can be printed in hexadecimal using the %x print format
Anyway, here is a single linked list, without keeping track of the tail of the list, or counting the elements.
#include <stdio.h>
#include <stdlib.h>
typedef struct listnode
{
int data;
struct listnode* next;
} linkedList;
linkedList* makeList(int a, int b, int c);
void addToList(linkedList* ll, int a);
void ListPrint(linkedList* ll);
int main()
{
linkedList* ll = makeList(1,3,5);
addToList(ll, 7);
addToList(ll, 9);
ListPrint(ll);
return 0;
}
linkedList* ListNew(int a) //new linkedList node
{
linkedList* newL = (linkedList*)malloc(sizeof(linkedList));
newL->data = a;
newL->next = NULL;
return newL;
}
linkedList* makeList(int a, int b, int c)
{
linkedList* ll = ListNew(a);
addToList(ll, b);
addToList(ll, c);
return ll;
}
void addToList(linkedList* ll, int a)
{
if(!ll) return;
//find end of list
while (ll->next)
{
ll = ll->next;
}
ll->next = ListNew(a);
return;
}
void ListPrint(linkedList* ll) //print list
{
if(!ll) return;
linkedList* p;
for( p=ll; p; p=p->next )
{
printf("%x: %d\n",p,p->data);
}
return;
}