C语言，我该如何打印指针-错误的输出

Question

I wrote an implementation of linked lists (with two points, one for the next value, and one for the previous value) in C language and I'm trying to test my code.

I checked that it scans and prints correctly, however, when I try to test the code I wrote to find a value in list, it returns incorrect output.

The code for find in list is:

node* find_in_list(node *anchor,data_type x)
{
    node *temp;
    int is_empty=0;
    is_empty=is_list_empty(anchor);
    if(is_empty)
        return NULL;
    temp=anchor->next;
    while(temp!=NULL)
    {
        if(temp->data==x)
            return temp;
        temp=temp->next;
    }
    return temp;
}

The code to check if the list is empty is

int is_list_empty(node *anchor)
{
    int boolean=0;
    if(anchor->next=NULL)
        boolean=1;
    return boolean;
}

It should be noted that anchor never changes. I define anchor as a node in the linked list that does not have an actual value, instead I just use it as a pointer to the first "real" node.

The void main is

#include "linked_list.h"
void main()
{
    data_type num;
    node *anchor;
    anchor=create_node();
    scan_list(anchor);
    printf("The list scanned is \n");
    print_list(anchor);
    printf("Enter the number you wish to find\n");
    scanf("%d",&num);
    printf("The address of %d is\n",num);
    printf("%p",find_in_list(anchor,num));
    getch();
}

The scanning and printing are done properly. It does indeed print the correct list, but when I try to print the address of some value in the list (no matter what value I enter) it returns 000000.

What is the problem?

Answer 1

I know you've already solved your problem, but ultimately a more straight forward algorithm may have prevented your issue in the first place.

From an elegance/beautiful/simplicity point-of-view, I'd re-write find_in_list() to be something like the following, after eliminating the is_list_empty() routine:

node* find_in_list(node *list,data_type x)
{
    for(;list;list=list->next)
        if(list->data==x)
          break;

    return list;
}

(edited to use for-loop)