Understanding how memory is allocated in driver code PSET 4, Filter (Less Comfortable)?

Question

What does the following line do, step by step (filter.c L:78)

RGBTRIPLE(*image)[width] = calloc(height, width * sizeof(RGBTRIPLE));

I am most confused by the syntax of

RGBTRIPLE(*image)[width]

What i believe I understand

• RGBTRIPLE is a struct
• image will be the variable name, which will be a pointer to a location in memory where there will be be the first element of an array of RGBTRIPLE's, which is the first location in the block allocated by calloc
• the array pointed to be image will be [width] elements long
• calloc is setting aside a block of memory initialized to zero, that is == the size of RGBTRIPLE's in a width x height image

What I am confused about

• What are the parentheses doing?
• Why are there no spaces?
• Why is width being used, and not height (isn't each element in the array a row in the image, not a column?)

Answer 1

RGBTRIPLE(*image)[width] means variable image is a pointer to an array of length width of RGBTRIPLE.

RGBTRIPLE *image[width] means variable image is an array of width pointers to RGBTRIPLE.

The parentheses couple the * closer to the variable, and change the meaning.

Spaces are often optional, style guides want them in a specific way or another.

width is used because of how those 2D arrays work.

It's width because that corresponds to the second index. image[i][j] will look at the memory address of image, add i times the size of RGBTRIPLE[width], and to the result will add j times the size of RGBTRIPLE.

#include <stdio.h>

int main() {
    int array[23][42];
    printf("size of array:   %zu\n", sizeof(  array));
    printf("size of row:     %zu\n", sizeof( *array));
    printf("size of element: %zu\n", sizeof(**array));
    return 0;
}

And for the fun of it, &array, &array[0], and &array[0][0] will report the same memory address.

Answer 2

What are the parentheses doing?

As Blauelf explained, they are changing order of operations. I didn't full grasp why this was needed though until trying to compile without them.

filter.c:78:15: error: variable-sized object may not be initialized
RGBTRIPLE*image[width] = calloc(height, width * sizeof(RGBTRIPLE));

They ensure *image creates a ptr called image, and [width] is treated separately as an array dimension. Still don't fully understand what is being attempted without them, but I now understand their necessity...

My main point of confusion was that I understood this works totally fine...

int n = 4;
int *d_arr = malloc(sizeof(int) * n);
for (int i = 0; i < n; i++)
{
  d_arr[i] = i;
  printf("1d: [%i] is %i\n", i, d_arr[i]);
}

So I assumed that this would also work fine...

  int n = 4;
  int *dd_arr = malloc(sizeof(int) * n * n); 
  for (int i = 0; i < n; i++)
  {
    for (int j = 0; j < n; j++)
    {   
      dd_arr[i][j] = i + j;
      printf("2d: [%i][%i] is %i\n", i, j, dd_arr[i][j]);
    }   
  }
}

But it does not, and won't compile with

2d_arr.c:12:16: error: subscripted value is not an array, pointer, or vector
      dd_arr[i][j] = i + j;
      ~~~~~~~~~^~
2d_arr.c:13:53: error: subscripted value is not an array, pointer, or vector
      printf("2d: [%i][%j] is %i\n", i, j, dd_arr[i][j]);

You have to have line two as

int (*dd_arr)[n] = malloc(sizeof(int) * n * n);

Which brings me to my understanding of the following...

Why is width being used, and not height (isn't each element in the array a row in the image, not a column?)

Best as I understand it, *dd_arr can be treated like an array out of the box, but since a 2d array is really an array OF arrays, you need to let the compiler know this so you can operate on that nested level (or another way, "image" is a pointer to an array that contains "width" number of arrays of RGBTRIPLES)

^^ Is that right?

And in the original question it's width not height, because height is the length of the outermost array implied by the *image, which doesn't need to be specified.

Why are there no spaces?

As Blauelf explained, not functional, only stylistic choice (albeit confusing to me at the time).