C, passing 2 dimensional array

Question

I haven\'t used pure C in a few years now, but I can\'t seem to make this really basic use case work. Here is the simple use-case in simple C, the actual situation is wrapp

Answer 1

void print_data(float **data, int I, int J)

expects an array of pointers to (the first element of arrays of) float.

But when you pass

float data[4][6];

you pass a pointer to float[6].

So in print_data, an access to

data[i]

reads sizeof(float*) bytes at an offset of i * sizeof(float*) bytes after what address data holds, and interprets these bytes as a float* that it then dereferences (after adding a suitable offset) in data[i][j].

So when you pass your 2D array, some float values are interpreted as pointers and then followed. That often leads to a segmentation fault.

You can either declare

void print_data(float (*data)[6], int I, int J)

and pass your 2D array, or you need to pass an array of pointers,

float *rows[4];
for(i = 0; i < 4; ++i) {
    rows[i] = &data[i][0];
}

and pass rows. Or, the third possibility is to pass and expect a flat array

void initialize_data(float* data, int I, int J) {
    for(i = 0; i < I; ++i) {
        for(j = 0; j < J; ++j) {
            data[i*J + j] = whatever;
        }
    }
}

and pass &data[0][0] from main.

Answer 2

A bi-dimensionnal array is not evaluated as a pointer to pointer, so you need to use an array of a pointer to array in your prototype:

void print_data(float data[4][6]);
void print_data(float (*data)[6]);