Question

What would the prototype declaration of a C function look like that returns an n-dimensional array?

Here, n>=2.

Please explain it with dynamic-array/pointers.

My idea is not to pass any-dimensional array.

Please give me 3 examples with 2, 3 and 4 dimensions.

That will be enough for me to grab the idea.

Answer 1

I would suggest this:

array foo();

where array is an struct defined as:

struct array
{
    int *data;
    unsigned int ndim;  //number of dimensions
    unsigned int *size; //size of each dimension is size[i].
};

Of course, data is just a pointer to int, but the other two fields in the struct can be used in such a way that the data can be interpreted like an n dimensional array, and size of each dimension can be stored in size which is yet another pointer.

For example, take 4-dimensional array (dimension size are 10, 20, 30 and 40), then you can create and initialize 4D array as:

unsigned int size[] = {10,20,30,40};
array arr = create(4, size);

where create function is defined as:

array create(unsigned int n, unsigned int *size)
{
    array arr;
    arr.ndim = n;
    arr.size = (unsigned int*) malloc(n * sizeof(unsigned int));
    int i;
    unsigned int totalElements =  1;
    for( i = 0 ; i < n ; ++i) 
    {
          arr.size[i] = size[i]; 
          totalElements *= size[i]; 
    }
    arr.data = (unsigned int*) malloc(totalElements * sizeof(int));
    return arr;
}

Of course, you’ve to work a lot with the fields of the struct, to make it look like n-dimensional array. You don’t need to follow me exactly the way I explained, but this is just a basic idea. You can modify it, to suit your specific need.

I would suggest you to write few functions, to manipulate the array, and to access the array elements. I’ve written create function, likewise here is destroy function:

void destroy(array arr)
{
    if (arr.size != NULL && arr.data != NULL) 
    {
       free(arr.size);
       free(arr.data);
       arr.size = arr.data = NULL;
    }
}