Numpy Like Arrays for C

Question

In numpy, you can access the size and shape of the array with array.size and array.shape while elements can be accessed using array[i].

How does one achieve this with a C structure? One can do something like

struct{
    int size;
    int shape[2];
    int *elements;
} int_array;

but then elements can be accessed as

int_array.elements[i].

not like numpy.

How can I have an array that stores size and shape, but whose elements can be accessed in the usual way using []?

Should be possible with appropriate (local) VLA typedef (and type cast or helper variable of that type). — hyde, Nov 06 '17 at 06:28
If C++ is an option, [check out `xtensor`](https://github.com/QuantStack/xtensor) — Eric, Nov 06 '17 at 06:29

score 2 · Answer 1 · 2017-11-06T06:13:19.967

2

Allocate an array that is 3 cells bigger than you need.
Put size and shape in the first 3 cells
Increment the pointer by 3 * sizeof(int).
Write accessors for size and shape.

You can then access the array in the usual way.

Don't forget to subtract 3 * sizeof(int) from the pointer before you free it.

edited Nov 06 '17 at 06:13

answered Nov 06 '17 at 05:51

well this works only directly for arrays where the element types are big enough to hold the dimensions... – Antti Haapala -- Слава Україні Nov 06 '17 at 07:51

score 1 · Answer 2 · answered Nov 06 '17 at 06:41

You could use a simple macro to cast your pointer to a 2D array (source):

#define INDEX(a) ((int(*)[a.shape[1]]) a.elements)

Which you can use as

#include <stdio.h>
#include <stdlib.h>

typedef struct{
    int size;
    int shape[2];
    int *elements;
} int_array;


int main() {
    int_array arr;
    arr.shape[0] = 2;
    arr.shape[1] = 3;
    arr.elements = malloc(6*sizeof(*arr.elements));

    INDEX(arr)[0][1] = 2;

    printf("%d", INDEX(arr)[0][1]);
}

Numpy Like Arrays for C

2 Answers2