Question

If this is a beginner’s question, my apologies – most of my programming has been in very high level langauges, and I have limited expertise in C. (This is the sort of thing I could do very easily in languages such as Matlab, Octave, Sage, Maxima etc, but for this I need the speed of C).

But anyway… I have an array whose size is set at run time with malloc:

int *A = malloc(m * sizeof(int));

where m is computed from some values provided by the user. I have a function “update” which updates the array (or, if you prefer, takes the array as input and returns another as output). This update function may be called upwards of 10^8 times.

So the function itself can’t introduce the appropriately sized output array with malloc, or the memory will be used up. So, for example, I can’t do this:

int * update(int *L) /* produces next iteration of L */
{
  int *out = malloc(m * sizeof(int));
  /* do some stuff with L and produce new array out */
  return (out);
}

I’ve tried to make out a static variable outside the update function:

static int *out;

and define its size in main:

out = malloc(m * sizeof(int));

But this doesn’t seem to work either.

Anyway, I would be very grateful of some advice – I think I’ve exhausted the excellence of google.

Answer 1

Allocate the array outside of update, then pass a pointer to it:

void update(int const *L, int *out)
{
    // whatever
}

Call as

int *A = malloc(m * sizeof(int));
if (A == NULL)
    // handle error

for (i=0; i < N_ITER; i++)
     update(L, A);

Though you may want to redesign the program so that it updates L in-place.