将指针的struct数组作为函数参数传递？

Question

The code works if i use the name of the struct array directly for the allocation, but not from the function argument. Otherwise it returns memory error.

typedef struct COORD
{
  int xp;
  int yp;
} coord;

coord** xy;

void allocate(coord** COORD)
{
  int i;

  //allocate COORD[500][460]
  COORD = (coord**)malloc(sizeof(coord*)*500);
  for(i=0; i<500; i++)
  {
    COORD[i] = (coord*)malloc(sizeof(coord)*460);
  }

  // freeing
  for (i=0; i<500; i++) free(COORD[i]);
  free(COORD);
}
//function call: allocate(xy);
//That is the code that leeds to the error

Using just xy instead of COORD works. And i am all wondering why is that not working.

Answer 1

You are mixing up various coding styles here. It's not clear what exactly you want to achieve. Pick one according to your task.

Temporary buffer

You need a large temporary buffer that should be allocated on the heap and that does not need to be seen from outside. Just create a local variable:

void do_stuff(int w, int h)
{
    coord **p;
    int i;

    p = malloc(h * sizeof(*p));
    for (i = 0; i < h; i++) p[i] = malloc(w * sizeof(**p));;

    // do stuff

    for (i = 0; i < h; i++) free(p[i]);
    free(p);
}

Allocate memory for further use

You want to allocate storage that your client code can use. Then provide two functions, one that allocates and one that frees the memory:

coord **create(int w, int h)
{
    coord **p;
    int i;

    p = malloc(h * sizeof(*p));
    for (i = 0; i < h; i++) p[i] = malloc(w * sizeof(**p));

    return p;
}

void destroy(coord **p, int h)
{
    int i;

    for (i = 0; i < h; i++) free(p[i]);
    free(p);
}

Your client code can then use the memory between these calls:

coord **p = create(500, 460);

// do stuff

drestroy(p, 500);

(Note that you have to pass the height to destroy , which is a bit unfortunate. It might be cleaner to create a wrapper struct that hold information about width and height and the pointer.)

Allocate memory for a global variable

You have a single instance of a global pointer. Then your functions always operate on that pointer and you don't need any further information on it (except the dimensions):

coord **global = NULL;

void destroy_global(int h)
{
    int i;

    for (i = 0; i < h; i++) free(global[i]);
    free(global);
    global = NULL;
}  

void create_global(int w, int h)
{
    int i;

    if (global != NULL) free_global();

    global = alloc(h * sizeof(*global));
    for (i = 0; i < h; i++) global[i] = malloc(w * sizeof(**global));        
}

Note that you should include <stdlib.h> for all memory functions and the NULL macro.

Addendum According to your comment, you want to allocate memory for a bitmap. That's option 2 above.

I recommend to create an object structure. You can pass a pointerv to that structure as handle to a bunch of functions. You can create the object with a function that returns that handle.

The following sketches a rough design for a bitmap object.

typedef struct Pixel Pixel;
typedef struct Bitmap Bitmap;

struct Pixel {
    uint8_t r, g, b;
};

struct Bitmap {
    int height;
    int width;
    Pixel **pixel;
};

Bitmap *bitmap_new(int w, int h)
{
    Bitmap *bmp = malloc(sizeof(*bmp));
    int i;

    bmp->height = h;
    bmp->width = w;
    bmp->pixel = malloc(h * sizeof(*bmp->pixel));
    for (i = 0; i < h; i++) {
        bmp->pixel[i] = malloc(w * sizeof(**bmp->pixel));
    }

    return p;
}

void bitmap_delete(Bitmap *bmp)
{
    int i;

    for (i = 0; i < h; i++) free(bmp->pixel[i]);
    free(bmp->pixel);
    free(bmp);
}



Bitmap *bitmap_read(const char *fn)
{
    Bitmap *bmp;
    FILE *f = fopen(fn, "rb");

    // read and allocate 
    return bmp;
}

void bitmap_blank(Bitmap *bmp, int r, int g, int b)
{
    for (i = 0; i < bitmap->height; i++) {
        for (j = 0; j < bitmap->width; j++) {
            bmp->pixel[i][j].r = r;
            bmp->pixel[i][j].g = g;
            bmp->pixel[i][j].b = b;
        }
    }
}

void bitmap_mirror_x(Bitmap *bmp)
{
    // do stuff
}

int bitmap_write(Bitmap *bmp, const char *fn)
{
    FILE *f = fopen(fn, "rb");

    // write bitmap to file
    return 0;
}

The design is similar to the interface to FILE * : fopen gives you a handle (or NULL ; error checking is omitted in the code above) and fread , fprintf , fseek and family take a pointer to the file as argument. Finally call fclose to close the file on disk and to free any ressources fopen has claimed.

Answer 2

The problem is that the function allocate() cannot change the value of xy outside itself. This is because C is call by value , the called function only gets the values of its arguments, not any kind of references to the expressions in the caller's context.

It needs to be:

void allocate(coord ***c)
{
}

and:

coord **xy;
allocate(&xy);

which of course is silly: the proper design would be for allocate() to return the new address:

coord ** allocate(void)
{
}

with use like:

coord **xy = allocate();

Probably it would be even better to have the dimensions as parameters to the function, since magic numbers are generally not a good thing:

coord ** allocate(size_t width, size_t height);

Answer 3

Have you tried to compile this code? There are a number of errors.

First, the type of main should always be 'int main(int argc, char *argv[])' Second, you need to '#include <stdlib.h>' at the top of your file to get the return type of malloc/free and friends. Third, you are not declaring 'i'. Fourth, you are using the same name 'COORD' as both a struct name and as a variable. Don't do this, it will cause you problems.

Sending incorrect code makes it very difficult to figure out what the root of your problem is, but I suspect it's the overloading of 'COORD'.

Answer 4

typedef struct COORD
{
  int xp;
  int yp;
} coord;

coord** xy;

void allocate(coord** COORD)
{
  int i;

  //allocate COORD[500][460]
  COORD = (coord**)malloc(sizeof(coord*)*500);
  for(i=0; i<500; i++)
  {
    COORD[i] = (coord*)malloc(sizeof(coord)*460);
  }

  // freeing
  for (i=0; i<500; i++) free(COORD[i]);
  free(COORD);
}
//function call: allocate();
//That is the code that works

Answer 5

typedef struct
{
  int xp;
  int yp;
} Coord;

Coord **xy;

Coord** allocate(size_t height, size_t width)
{
  int i;
  Coord **arr;

  arr = malloc(sizeof(Coord*)*height);
  for(i=0; i<height; i++) {
    arr[i] = malloc(sizeof(coord)*width);
  }
  return arr;
}

void allocate2(Coord ***p_arr, size_t height, size_t width)
{
  int i;
  Coord **arr;

  arr = *p_arr;

  arr = malloc(sizeof(Coord*)*height);
  for(i=0; i<height; i++) {
    arr[i] = malloc(sizeof(coord)*width);
  }
}

void deallocate(Coord **arr, size_t height)
{
  for (i=0; i<500; i++) {
    free(arr[i]);
  }
  free(arr);
}

int main()
{
  Coord **arr_2;
  Coord ***p_arr_3;

  allocate2(&xy, 500, 460);
  /* do something with global array, xy, e.g. */
  xy[1][2].xp = 100;
  xy[1][2].yp = 200;
  deallocate(xy, 500);

  arr_2 = allocate(500, 460);
  /* do something with local array, arr_2 */
  deallocate(arr_2, 500);

  allocate2(p_arr_3, 500, 460);
  /* do something with ptr to local array, p_arr_3 */
  deallocate(*p_arr_3, 500);

  return 0;
}