? pointer

bsd_lite

Hi all,

I found the following function, which is used to remove the head of a
linked list.

void RemoveHead(node **head)
{
    node *temp;
    if (head && *head) { /* Corrected code */
        temp = (*head)->next;
        free(*head);
        *head = temp;
    }

}

1. Why is the purpose of 'head' in the 'if' condition? Shouldn't
'*head' suffice?

2. This function has an argument, which is a pointer to a pointer.
Presumably, 'head' is a pointer to '*head', which is another pointer
to the address of the link list head. If 'head' does not point to
NULL, does it follow that '*head' does not point to NULL? If so, why?

tyc611

1. 仅当head是一个有效的指针时，才可以解引用该指针(*head)
2. 不能推出。head指向一个指针变量，它的值不为NULL，但它指向的指针变量可以为任何值（即包括NULL）

bsd_lite

oh, thank you.
as the following example?

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

  static void showtext(const char **ptr);

  int
  main(void)
  {
      const char *text = NULL;
      const char **ptr = &text;

      assert(ptr != NULL);
      showtext(ptr);

      text = "Hello world";
      showtext(ptr);

      return EXIT_SUCCESS;
  }

  static void
  showtext(const char **ptr)
  {
      if (ptr == NULL)
          printf("invalid argument\n");
      if (*ptr == NULL)
          printf("a null pointer\n");
      else
          printf("a pointer to the text `%s'\n", *ptr);
  }

bsd_lite

$ ./a.out
  a null pointer
  a pointer to the text `Hello world'
  $

tyc611

you got it