   Introduction to Pointers, Structures and Linked-Lists Part  3
  Comments for the article Introduction to Pointers, Structures and Linked-Lists Part  3
  [in reply to the entire series, not just this part]Thank you dwarfsoft!  I can honestly say this is probably the best linked-list tutorial I've read and it really helped me get ahold of some of the ideas I just wasn't grasping.  I found the article so good, in fact, that by the time I got done coding the final details of my happy little linked list, I had completely forgotten that I didn't really need one..So anyway, thank you and good job :)   ;  I have been having lots of trouble with too much code lately in c++, and i found this error (which i assume is something i've done wrong)This piece of code:GridSquare_t Square1,Square2;Square2.pNextSquare = NULL; Square1.pNextSquare = &Square2brings up this error:error C2440: '=' : cannot convert from 'GridSquare_t *__w64 ' to 'GridSquarePrt_t *'when working with the struct as follows:typedef struct sGridSquare *GridSquarePrt_t;typedef struct sGridSquare{int X;int Y;char Z[20];struct GridSquarePrt_t *pNextSquare;} GridSquare_t;;  my guess is thattypedef struct sGridSquare *GridSquarePrt_t;typedef struct sGridSquare{int X;int Y;char Z[20];struct GridSquarePrt_t *pNextSquare;} GridSquare_t;pNextSquare is a pointer to a struct sGridSquarePtr_t, which is a pointer to a sGridSquareit looks more like this to the compilerstruct sGridSquare{int X;int Y;char Z[20];struct sGridSquare **pNextSquare; // a pointer to a pointer};just get rid of the asterix in the declaration of pNextSquare   