Memory allocation in D (noob question)

[Steven Schveighoffer]

"Oskar Linde" wrote
> mandel wrote:
>> Steven Schveighoffer wrote:
>> [..]
>>> Now I create the valid array slices:
>>>
>>> int[] array3 = array1[$..$];
>>> int[] array4 = array2[0..0];
>>>
>>> Note that both of these arrays are bit-for-bit identical (both have 0
>>> length and the same ptr value).  Which one points to which piece of
>>> memory?  How is the GC to decide which memory gets collected?
>> I see the problem.
>> The first possible solution that comes to my mind seeing this is to
>> make array1[0..0] and array1[$..$] equal.
>> array1[$..$] could point to the begin of the array.
>> Since the slice length is null, it shouldn't matter - would it?
>
> Appending to a (empty or not) array slice starting at the start of an 
> allocated block appends in-place rather than allocate a new array. This is 
> the reason
>
> while(x)
>   a ~= b;
>
> can be reasonably efficient.

Hm... I think you are slightly incorrect.  I think the array is appended to 
in place ONLY if the data after the slice is unallocated.  In this case, it 
would be allocated, so the array would be re-allocated elsewhere.

> So appending to the [$..$] array would (without padding) mean that you 
> corrupt the following array.

I think this is incorrect for the reasons I stated above.  An allocated 
block should never be re-assigned to another array.

Maybe I am wrong, but I think mandel might have a possible solution to this 
problem.  If you slice an empty array (or even allocate an empty array), set 
the pointer to null.  No reason to allocate an empty array, and no reason 
you need to keep memory around for it.  If you append to it, it's going to 
be like appending to an init array anyways.  That would also make null 
comparisons more consistent like:

int[] array1 = array2[0..0];

if(array1 is null) // evaluates to true!
...

-Steve