Question

我有一个结构：

 struct room;
 ...
 /*some more structs */

 typedef struct {     
   int state;
   int id;
   struct room* north;
   struct room* south;
   struct room* east;
   struct room* west;
   creature* creatures[10];
 } room;

打印内容的函数：

 void printContentsOfRoom(room* r) {
   printf("\nRoom %d:\n", r->id);
   printf("\tState: %s\n", getState(r));
   printf("\n\tNeighbors:\n");
   if (r->north.id != -1)
     printf("\t    North: Room %d\n",((room*)r->north)->id);
   if (r->east.id != -1)
     printf("\t    East: Room %d\n",((room*)r->east)->id);
   if (r->south.id != -1)
     printf("\t    South: Room %d\n",((room*)r->south)->id);
   if (r->west.id != -1)
     printf("\t    West: Room %d\n",((room*)r->west)->id);
   printf("\n\tCreatures:\n");
   for (int i = 0; i < 10; i++) {
     creature* c = (creature*)r->creatures[i];
     if (c) {
       if (c == pc)
     printf("\t    %s\n",getType(c));
       else
     printf("\t    %s %d\n",getType(c),c->id);
     }
   }
 }

我试图这样做，以便如果房间ID设置为-1（在程序运行时设置），它不会打印有关该房间的信息。在编译时，我收到错误＆＃34;请求成员＆＃39; id＆＃39;在某种结构或联合的东西中。＆＃34;

此外，我尝试设置if条件r-＆gt; north-＆gt; id，并返回错误＆＃34;解除引用指向不完整类型＆＃39; struct room＆＃39;＆＃34; < / p>

Answer 1

我认为你想做这样的前瞻声明：

typedef struct room room;

struct room{
    .
    .
    .
};

然后您可以使用r->north->id。请参阅此问题：How to define a typedef struct containing pointers to itself?

如何从指针的指针引用变量中的值？

1 个答案: