C语言 如何在C中使用指向结构的指针从结构中打印成员数据

How to print member data from a struct using pointer to struct in C

提问by Bradford

I have a pointer to a struct of type Map defined in an external header file:

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;

The pointer is initialised as follows:

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.

How do I go about printing the values of width and height stored within the Map structure which is created in create_map? create_map is held in an external file and the only variable it passes back to main is the pointer to the map.

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 保存在一个外部文件中，它传递回 main 的唯一变量是指向映射的指针。

The following gives an error when compiling ("error: dereferencing pointer to incomplete type")

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);

As far as I know, the pointer is valid as the code below prints a memory address:

据我所知，该指针是有效的，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);

回答by P.P

Just drop the structkeyword from:

只需struct从以下位置删除关键字：

struct Map *map_ptr;    

to:

到：

Map *map_ptr;    

You have declared a nameless struct and typedef'ed it to Map. So when you declare struct Map *map_ptr;, compiler thinks this is another structcalled Map.

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.

回答by Jens

You tripped over what is called namespacesin C. There are separate namespaces for

你被C 中所谓的命名空间绊倒了。 有单独的命名空间

• typedef names, as you introduced with typedef struct { ... } Map;
• struct tags, as you introduced with struct Map *map_ptr;
• plus other namespaces for objects, macro names, ...

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

The same indentifier can be reused in different namespaces. I recommend to never bother with typedefs for structs. It only hides useful information, and all it does it saving you from writing structevery now and then. If something is a struct or pointer to a struct then I want to know it so I know whether to use ->or .to access the members. Using typedefs defeats this by hiding useful information.

相同的标识符可以在不同的命名空间中重复使用。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您免于时不时地写作struct。如果某个东西是结构体或指向结构体的指针，那么我想知道它，以便我知道是使用->还是.访问成员。使用 typedef 可以通过隐藏有用的信息来解决这个问题。

One way to fix your problem is to get rid of the typedef and only use a struct tag with

解决问题的一种方法是去掉 typedef 并只使用带有

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 = ...;

回答by paulsm4

Here is a complete example that might help clarify a few points:

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;
} 

EXAMPLE OUTPUT:

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...

An alternative approach would be to use "struct Map {...}" instead of the typedef:

另一种方法是使用“struct Map {...}”而不是typedef：

EXAMPLE:

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);

回答by white

Answer 1：

回答 1：

struct Map *map_ptr; 

to

到

Map *map_ptr; 

Answer 2：

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;

to

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.
} ;

Reason ：

原因 ：

if typedef struct{...}  B； 

so

所以

B ＝＝ struct B{...}