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

Question

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

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:

    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.

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

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

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

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

How does it not work? Does it not do anything? Does it crash? What happens? — icktoofay
– icktoofay, Commented Dec 30, 2012 at 22:19
C nit: %p args must be (cast to) ptr-to-void otherwise you invoke undefined behavior. — Jens
– Jens, Commented Dec 30, 2012 at 23:03
See also this link, or the sample code below: publib.boulder.ibm.com/infocenter/comphelp/v8v101/… — paulsm4
– paulsm4, Commented Dec 31, 2012 at 1:28

P.P · Accepted Answer · 2012-12-30 22:30:09Z

5

Just drop the struct keyword from:

struct Map *map_ptr;

to:

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 struct called Map.

answered Dec 30, 2012 at 22:30

P.P

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Comments

Jens · Accepted Answer · 2012-12-30 22:59:33Z

2

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

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, ...

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 struct every 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.

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

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 = ...;

answered Dec 30, 2012 at 22:59

Jens

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1 Comment

paulsm4 Over a year ago

Very, very good point - thank you. I should emphasize that the C concept of "namespace" does not correspond to C++ "namespaces". The two are different. Here's a link: publib.boulder.ibm.com/infocenter/comphelp/v8v101/…

paulsm4 · Accepted Answer · 2012-12-30 22:36:37Z

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

#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:

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:

EXAMPLE:

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);

Steve Vinoski · Accepted Answer · 2015-09-21 03:12:49Z

0

Answer 1：

struct Map *map_ptr;

to

Map *map_ptr;

Answer 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

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 ：

if typedef struct{...}  B；

so

B ＝＝ struct B{...}

edited Sep 21, 2015 at 3:12

Steve Vinoski

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answered Dec 31, 2012 at 3:18

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