Correct way to allocate and free arrays of pointers to arrays

Question

I want to create an array of pointers to arrays of 3 floats. What is the correct way to do this?

float *array1[SIZE]; // I think it is automatically allocate         


        

Answer 1

A general rule is that for each time you call malloc() or calloc() you will need to do a free() call on the returned pointer.

If you want a two dimensional array with compile-time known size, just use a two dimensional array! float val[5][3] is perfectly valid.

If you want a two dimensional array and you don't know it's size during compile-time, you most probably want to use a standard, single diemensional calloc() and an appropriate getter.

#define ARR_COLUMNS 10
#define ARR_ROWS 10
float* arr = calloc (ARR_COLUMNS * ARR_ROWS, sizeof(float));

int get(float* arr, int x, int y) {
  if (x<0 || x>= ARR_COLUMNS) return 0;
  if (y<0 || y>= ARR_ROWS) return 0;
  return arr[ARR_COLUMNS*y+x];
}

void set (int* arr, int x, int y, float val) {
  if (x<0 || x>= ARR_COLUMNS) return;
  if (y<0 || y>= ARR_ROWS) return;
  arr[ARR_COLUMNS*y+x] = val;
}

Of course replace the defines with appropriate variables.

By doing so you will:

• save yourself costly allocs and frees
• have less fragmented memory
• simplify your possible realloc calls
• ensure the data is cached better and accessed without the common [x][y] vs [y][x] iteration cache problem.