simple linked list

I'm using the following short code to help me understand linked lists (keep things simple at this stage!). The code compiles with the warning -
": warning C4133: "=" : incompatible types - from 'struct RECORD *' to 'struct RECORD *' .

All I am checking with this code is that I can assign the address of the next location in the list to the current record (BTW I know that I have only one record in the list ).

When I run the program it appears to work ok, but that warning is confusing me - can anyone help?

Regards

Code:

---

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

#define listSize 10



/* global structure definitions */
typedef struct
                {
                  char                        name[10];
                  struct RECORD        *nextPtr;
                       
                } RECORD; /* end structure */


int main(void)
{
/* declare local variables and initialise  */
    RECORD myRecord = {"abc", NULL};
    RECORD *listPtrStart = NULL, *listPtrCurrent = NULL;



    /* allocate sufficient memory to store the list */
    listPtrStart = (RECORD *) malloc(listSize*sizeof (RECORD));

    if(listPtrStart != NULL)
    {
            listPtrCurrent = listPtrStart;

            /* insert a copy of 'myRecord' into the list */
        *listPtrCurrent = myRecord;

            /* move list pointer to next location and store its value */
        listPtrCurrent->nextPtr = listPtrCurrent + 1;
               
      }
      else
      {
                printf("Memory not allocated");
      }



      return 0; /* indicate successful termination */


}

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Code:

---

typedef struct RECORD

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Hi there. I'm a bit new to the forums but I think the above poster meant that your structure should be declared:

Code:

---

typedef struct _RECORD
{
        char                        name[10];
        struct _RECORD        *nextPtr;
                       
} RECORD;

---

Since you're using the struct you declared within it's own definition.

You also don't need to have a macro for the list's size because linked lists by nature have no maximum size. So you're allocating way too much space with that malloc. I think it would be better to allocate each node (RECORD struct) as you go along and I wrote up an example of what I'm talking about:

Code:

---

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

typedef struct _RECORD
{
  char                  *name;
  struct _RECORD        *nextPtr;
 
} RECORD;


int main(void)
{
  /* Allocating a new linked list (LL) */
  RECORD *myRecord = (RECORD *) malloc( sizeof( RECORD ) );
  if (!myRecord) { printf("Memory not allocated\n");  return -1; }

  /* Couple of strings to add to LL's later */
  char *str1 = "abc";
  char *str2 = "def";

  /* Setting up the LL's data members */
  myRecord->name = str1;
  myRecord->nextPtr = NULL;

  /* Allocating another LL to add */
  RECORD *newRecord = (RECORD *) malloc( sizeof( RECORD ) );
  if (!newRecord) { printf("Memory not allocated\n");  return -1; }
 
  newRecord->name = str2;
  newRecord->nextPtr = NULL;

  /* Now we "link" the first node to the 2nd node */
  myRecord->nextPtr = newRecord;

  printf("First node's str: %s\n", myRecord->name);
  printf("Successor's str: %s\n", myRecord->nextPtr->name);

  free(myRecord);
  free(newRecord);
  return 0;
}

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So the output of that code would be:
abc
def

I hope this helps and if anyone sees anything that I did wrong please let me know