Question

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的唯一变量是指向地图的指针。

以下编译时出错（“错误：解除指向不完整类型的指针”）

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

据我所知，指针有效，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);

Answer 1

只需删除struct关键字：

struct Map *map_ptr;

为：

Map *map_ptr;

您已声明了无名结构，并将其定义为Map。因此，当您声明struct Map *map_ptr;时，编译器认为这是另一个名为Map的结构。

Answer 2

你在C中绊倒了所谓的名称空间。

有单独的名称空间

typedef names，正如您在typedef struct { ... } Map;
struct tags，正如您在struct Map *map_ptr;
加上对象，宏名称等的其他命名空间......

可以在不同的命名空间中重用相同的标识符。我建议永远不要打扰结构的typedef 。它只隐藏有用的信息，所有这些都可以帮助您不时地编写struct。如果某个东西是结构或指向结构的指针，那么我想知道它，所以我知道是否使用->或.来访问成员。使用typedef会隐藏有用信息，从而使其失败。

解决问题的一种方法是摆脱typedef，只使用带

的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 = ...;

Answer 3

以下是一个完整的示例，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：

实施例

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);

Answer 4

答案1：

struct Map *map_ptr;

到

Map *map_ptr;

答案2：

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;

到

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：

if typedef struct{...}  B；

所以

B ＝＝ struct B{...}

如何使用指向C中struct的指针从结构中打印成员数据

4 个答案: