Pass a 2d numpy array to c using ctypes

Question

What is the correct way to pass a numpy 2d - array to a c function using ctypes ? My current approach so far (leads to a segfault):

C code :

void te         


        

Answer 1

#include 

void test(double (*in_array)[3], int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j < N; j++){
            printf("%e \t", in_array[i][j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };

    test(a, 3);
    return 0;
}

if you want to use a double ** in your function, you must pass an array of pointer to double (not a 2d array):

#include 

void test(double **in_array, int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j< N; j++){
            printf("%e \t", in_array[i][j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };
    double *p[] = {a[0], a[1], a[2]};

    test(p, 3);
    return 0;
}

Another (as suggested by @eryksun): pass a single pointer and do some arithmetic to get the index:

#include 

void test(double *in_array, int N){
    int i, j;

    for(i = 0; i < N; i++){
        for(j = 0; j< N; j++){
            printf("%e \t", in_array[i * N + j]);
        }
        printf("\n");
    }
}

int main(void)
{
    double a[][3] = {
        {1., 2., 3.},
        {4., 5., 6.},
        {7., 8., 9.},
    };

    test(a[0], 3);
    return 0;
}