C How to free nodes in the linked list

Question

How will I free the nodes allocated in another function   struct node       int data      struct node  next      struct node  buildList         struct node  head   NULL      struct node  second   NULL      struct node  third   NULL       head   malloc sizeof struct node        second   malloc sizeof struct node        third   malloc sizeof struct node         head- gt data   1      head- gt next   second       second- gt data   2      second- gt next   third       third- gt data   3      third- gt next   NULL       return head        I call the buildList function in the main      int main         struct node  h   buildList        printf  The second element is  d n   h- gt next- gt data       return 0        I want to free head  second and third variables  Thanks   Update     int main         struct node  h   buildList        printf  The element is  d n   h- gt next- gt data      prints 2       free h- gt next- gt next         free h- gt next       free h          struct node  h1   buildList        printf  The element is  d n   h- gt next- gt data      print 2    why      return 0      Both prints 2  Shouldn t calling free h  remove h  If so why is that h- next- data available  if h is free  Ofcourse the  second  node is not freed  But since head is removed  it should be able to reference the next element  What s the mistake here

User · Answer

Simply by iterating over the list   struct node  n   head  while n      struct node  n1   n     n   n- gt next     free n1

User · Answer

One function can do the job   void free list node  pHead        node  pNode   pHead   pNext       while  NULL    pNode                pNext   pNode- gt next          free pNode           pNode   pNext

User · Answer

An iterative function to free your list   void freeList struct node  head       struct node  tmp      while  head    NULL               tmp   head         head   head- gt next         free tmp              What the function is doing is the follow    check if head is NULL  if yes the list is empty and we just return Save the head in a tmp variable  and make head point to the next node on your list  this is done in head   head- gt next Now we can safely free tmp  variable  and head just points to the rest of the list  go back to step 1

User · Answer

struct node      int position      char name 30       struct node   next      void free list node   list       node  next node       printf   n n Freeing List   n        while list    NULL                next node   list- gt next          printf  clear mem for   s  list- gt name           free list           list   next node          printf  - gt

User · Answer

You could always do it recursively like so   void freeList struct node  currentNode        if currentNode- gt next  freeList currentNode- gt next       free currentNode

[c] C: How to free nodes in the linked list?

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