How to return a 2D array from a function in C?

Question

I am a Ruby programmer who has ended up developing a code generate for C. Its like asking a Limo to tow a 1960s truck. Any way.

Here is what I thought should work but doesnt work.

float[][] pixels()
{
  float x[][]= { {1,1},{2,2} };
  return x
}

void drawLine(float x[][2])
{
  //drawing the line
}

//inside main
drawLine(pixels());

I have banged my head on my desk trying to get this thing work. Please help.

Answer 1

In C, pointers and arrays are closely related. Also, you usually need to pass the size of an array as a separate variable. Let's start you with:

#include <stdio.h>

float** createArray(int m, int n)
{
    float* values = calloc(m*n, sizeof(float));
    float** rows = malloc(m*sizeof(float*));
    for (int i=0; i<m; ++i)
    {
        rows[i] = values + i*n;
    }
    return rows;
}

void destroyArray(float** arr)
{
    free(*arr);
    free(arr);
}

void drawLine(const float** coords, int m, int n);

int main(void)
{
    float** arr = createArray(2,2);
    arr[0][0] = 1;
    arr[0][1] = 1;
    arr[1][0] = 2;
    arr[1][1] = 2;
    drawLine(arr, 2, 2); 
    destroyArray(arr);
}

Answer 2

Thank you all for your answers and more specifically for the detailed explanation of the array-pointer relationship.

I encapsulated the array in a structure

 struct point_group1 {
        float x[3];
        float y[3];
};

struct point_group1 pixels(){
    struct point_group1 temp;

    temp.x[0] = 0.0;
    temp.x[1] = 1.0;
    temp.x[2] = -1.0;

    temp.y[0] = 0.0;
    temp.y[1] = 1.0;
    temp.y[2] = 1.0;

    return temp;    
}



struct point_group1 points1  = pixels();
axPoly(points1.x, points1.y ,3, 0.0);

Answer 3

float (*pixels(void))[2] 
{
  static float x[2][2]= { {1,1},{2,2} };
  return x;
}

void drawLine(float (*x)[2])
{
  //drawing the line
  //x[0][0];
}

//inside main
drawLine(pixels());

Answer 4

In C/C++, when you pass an array to a function, it decays to be a pointer pointing to first element of the array. So, in pixels() function, you are returning the address of a stack allocated variable. The returning variable's address is no longer valid because on pixels() return, the stack allocated variable goes out of scope. So, instead you should for a variable whose storage is dynamic ( i.e., using malloc, calloc ).

So, for a two dimensional array, you may use float** arrayVariable;. Also, if you passing this to a function, you should be wary of how many rows & columns it has.

int rows, columns;

float** pixels()
{
    // take input for rows, columns
    // allocate memory from free store for the 2D array accordingly
    // return the array
}

void drawLine( float** returnedArrayVariable )
{
  //drawing the line
}

Since, 2D array is managing resources it self, it should return the resources back to the free store using free.

Answer 5

The easiest way is probably going to be declaring the float array in main and having pixels just fill it:

#define PIXEL_X_SIZE 2
#define PIXEL_Y_SIZE 2

int pixels(float x[][PIXEL_X_SIZE], int len) {
    /* I don't know if you want the logic of this method to ever change,
       but this will be roughly equivalent to what you do above */
    if (len < PIXEL_Y_SIZE) {
        /* the length of the passed array is too small, abort */
        return -1;
    }

    x[0][0] = x[0][1] = 1;
    x[1][0] = x[1][1] = 2;
    return 0;
}

void drawLine(float x[][PIXEL_X_SIZE]) {
    /* this will work fine */
}

int main() {
    float pixel_array[PIXEL_Y_SIZE][PIXEL_X_SIZE];
    pixels(pixel_array, PIXEL_Y_SIZE);
    drawLine(pixel_array);
}

You can also use malloc and free and store your pixels on the heap, but if this is all the bigger the pixels array is going to be, there's really no need and it just adds additional complexity to make sure your memory always get properly allocated and freed.