2

我有一个指向外部头文件中定义的 Map 类型结构的指针：

typedef struct {
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
} Map;

指针初始化如下：

    struct Map *map_ptr;    
    map_ptr = create_map(*w_ptr, *h_ptr);
    // create_map returns Map*, w_ptr and h_ptr are pointers to height and width fields for a map/maze.

如何打印存储在 create_map 中创建的 Map 结构中的宽度和高度值？create_map 保存在外部文件中，它传递回 main 的唯一变量是指向映射的指针。

编译时出现以下错误（“错误：取消引用指向不完整类型的指针”）

printf("Height = %d\n", map_ptr->height);

据我所知，指针是有效的，因为下面的代码打印了一个内存地址：

printf("Pointer address for map = %p\n", map_ptr);
4

4 回答 4

4

只需从以下位置删除struct关键字：

struct Map *map_ptr;

至：

Map *map_ptr;

您已经声明了一个无名结构并将其类型定义为Map. 因此，当您声明时struct Map *map_ptr;，编译器认为这是另一个名为Map.

于 2012-12-30T22:30:09.603 回答
2

你绊倒了 C 中所谓的命名空间。有单独的命名空间

  • typedef 名称，正如您介绍的那样typedef struct { ... } Map;
  • struct 标签，正如您介绍的那样struct Map *map_ptr;
  • 加上对象的其他命名空间，宏名称，......

相同的标识符可以在不同的命名空间中重用。我建议永远不要为 structs 使用 typedef。它只隐藏有用的信息，它所做的一切都是为了让你不时地写作struct。如果某物是结构或指向结构的指针，那么我想知道它，以便知道是使用->还是.访问成员。使用 typedefs 通过隐藏有用的信息来解决这个问题。

解决问题的一种方法是摆脱 typedef 并仅使用 struct 标签

struct Map {
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
};
struct Map *map_ptr = ...;
于 2012-12-30T22:59:33.827 回答
0

答案一：</p> 

struct Map *map_ptr;

至

Map *map_ptr;

答案2：</p> 

typedef struct {
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
} Map;

至

struct Map{
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
} ;

原因：</p> 

if typedef struct{...}  B；

所以

B ＝＝ struct B{...}
于 2012-12-31T03:18:43.430 回答
0

这是一个完整的示例，可能有助于澄清几点：

#include <stdio.h>
#include <malloc.h>
#include <string.h>

typedef struct {
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
} Map;


Map *
create_map ()
{
  printf ("Allocating %d bytes for map_ptr, and %d bytes for map data...\n",
    sizeof (Map), 100);
  Map *tmp = (Map *)malloc(sizeof (Map));
  tmp->squares = (char *)malloc (100);
  strcpy (tmp->squares, "Map data...");
  tmp->width = 50;
  tmp->height = 100;
  return tmp;
}

int 
main(int argc, char *argv[])
{
  Map *map_ptr = create_map();
  printf ("map_ptr->height= %d, width=%d, squares=%s\n",
    map_ptr->height, map_ptr->width, map_ptr->squares);
  free (map_ptr->squares);
  free (map_ptr);
  return 0;
}

示例输出：

Allocating 12 bytes for map_ptr, and 100 bytes for map data...
map_ptr->height= 100, width=50, squares=Map data...

另一种方法是使用“struct Map {...}”而不是 typedef：

例子：

struct Map {
    char *squares; //!< A pointer to a block of memory to hold the map.
    int   width;   //!< The width of the map pointed to by squares.
    int   height;  //!< The height of the map pointed to by squares.
} Map;


struct Map *
create_map ()
{
...
  struct Map *tmp = (struct Map *)malloc(sizeof (struct Map));
  ...
}
  ...
  struct Map *map_ptr = create_map();
  printf ("map_ptr->height= %d, width=%d, squares=%s\n",
    map_ptr->height, map_ptr->width, map_ptr->squares);
  free (map_ptr->squares);
  free (map_ptr);
于 2012-12-30T22:29:23.387 回答