On Sun, 03 Apr 2011 08:29:47 -0400, simendsjo <[email protected]>
wrote:
D will copy the original content if a slice expands, but is the
following behavior
implementation specific, or part of the specification?
auto a = [0,1,2];
auto b = a[0..2];
a.length = 2; // is a[3] already marked for collection by the gc?
assert(a.ptr == b.ptr);
b.length += 1; // as b cannot reuse it even though noone is using it
assert(a.ptr == b.ptr); // ooops, fails.. b is reallocated
There are two incorrect assumptions here:
1. The D array runtime code cannot possibly know how many references there
are to a certain memory block, so your assertion that a[3] is unreferenced
is not something that the compiler or runtime can know without horrendous
performance implications.
2. The D GC collects entire memory blocks, not portions of them. There is
no way for the GC to "collect" a[3]. It can only collect the entire
memory block which contains a.
The array runtime code does it's best effort to expand an array into its
own memory block. Because slicing an array or copying an array reference
is a very fast, efficient event, it is not hooked by the array runtime
code. Therefore, the array runtime code cannot know all the array slices
that point to the memory block. The assumption it MUST make is that
somebody somewhere has a pointer to the data it considers valid. If you
think about it, the bookkeeping necessary to track how many
pointers/arrays point to each byte of an array would dwarf the cost of
storing and using the array in the first place.
You must remember, copy-on-expansion (note, copy-on-write is not the right
term, that implies changing any data in the array makes a copy), or rather
the lack of copying, is an optimization. In most cases, you should expect
expansion can make a copy, but you shouldn't rely on that behavior.
For those cases where you want to *force* the runtime to consider data not
referenced by an array as being 'free', you can use assumeSafeAppend:
b.assumeSafeAppend(); // marks b[3] as being unused
b.length += 1; // no copy made.
-Steve
