Behaviour change for double-free'ing a pointer

[Tim Cussins]

Hi Joel

On Nov 26th there was a commit that changed the behaviour of free() when
attempting to free a previously free'd pointer. Before the change, the
behaviour was to assert(0) - the new behaviour is to printk() a warning
and continue.

void free(
  void *ptr
)
{

  ...

  if ( !_Protected_heap_Free( &RTEMS_Malloc_Heap, ptr ) ) {
    printk( "Program heap: free of bad pointer %p -- range %p - %p \n",
      ptr,
      RTEMS_Malloc_Heap.start,
      RTEMS_Malloc_Heap.end
    );
}

IMHO the assert method is more correct, as an obvious programmer error
has been detected and can be caught and a debugger may be attached or
whatnot.

I guess there's a good argument for tolerating a double-free - primarily
that the second free is unlikely to cause disasterous software
behaviour. Contrast this with assert(), which is undeniably a total
disaster from a user perspective... :P

Perhaps the choice of behaviour could be based on RTEMS_HEAP_DEBUG?

Regards,
Tim