Before looking at the answer, try writing a simple C program (with a for loop) to do this. Quite a few people get this wrong.
This is the wrong way to do it
struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = listptr->next)
{
free(listptr);
}
If you are thinking why the above piece of code is wrong, note that once you free the listptr node, you cannot do something like listptr = listptr->next!. Since listptr is already freed, using it to get listptr->next is illegal and can cause unpredictable results!
This is the right way to do it
struct list *listptr, *nextptr;
for(listptr = head; listptr != NULL; listptr = nextptr)
{
nextptr = listptr->next;
free(listptr);
}
head = NULL;
After doing this, make sure you also set the head pointer to NULL!
Thursday, April 8, 2010
C program to free the nodes of a linked list
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