我不明白为什么下面的代码返回一个指针0x1。股票节点被发现是正常的，如下面的输出所示：

ticket_name passed comparison.
ticket_type passed comparison.
ticket_zone passed comparison.

但是返回的指针是0x1，我试图将该指针与链表中的指针进行比较以删除一个节点，这显然是失败的。在编译函数时也有一个警告：

tm_options.c:85: warning: assignment makes pointer from integer without a cast
tm_options.c:190: warning: assignment makes pointer from integer without a cast

据我所知，这个警告是因为函数缺少函数原型，而没有原型的函数的默认返回类型是int。

**编辑：**这里是我的Makefile的粘贴箱，以防万一，这与它有关http://pastebin.com/UFw9B4Hd

struct stock_node * find_ticket(tm_type_ptr tm, char *, char, char *);

下面是返回错误值的函数：

struct stock_node * find_ticket(tm_type * tm, char * ticketName, char ticketType, char * ticketZone) {

  struct stock_node * curr;
  BOOLEAN found = FALSE;

  curr = tm->stock->head_stock;

  while (curr != NULL && found == FALSE) {

    printf("curr: %p\n", (void *) curr);
    printf("curr->ticket_name: %s\n", curr->data->ticket_name);
    printf("curr->ticket_type: %c\n", curr->data->ticket_type);
    printf("curr->ticket_zone: %s\n", curr->data->ticket_zone);

      if (strncmp((char *)curr->data->ticket_name, ticketName, TICKET_NAME_LEN + 1) == 0) {

          printf("ticket_name passed comparison.\n");

          if ((char) curr->data->ticket_type == ticketType) {

              printf("ticket_type passed comparison.\n");

              if (strncmp((char *)curr->data->ticket_zone, ticketZone, TICKET_ZONE_LEN + 1) == 0) {

        printf("ticket_zone passed comparison\n");

        found = TRUE;
                return curr;

          }

       }

    } else {

       curr = curr->next_node;

    }

 }

 return NULL;

}

任何帮助将不胜感激。如果您需要更多的信息，请不要犹豫问我。
调用代码为：

struct stock_node * stockNode;
((stockNode = find_ticket(tm, ticketName, ticketType, ticketZone) != NULL))

根据要求：struct stock_node定义：

typedef struct stock_node
{
    struct stock_data * data;
    struct stock_node * next_node; 
} stock_node;

此外，tm_type_ptr：

typedef struct tm * tm_type_ptr;