On Wednesday, 20 July 2016 at 08:18:55 UTC, celavek wrote:
As far as my current understanding goes the shuffle will be
done in place.
If I use the "representation" would that still hold, that is
will I be able
to use the same char[] but in the shuffled form? (of course I
will test that)
representation does not allocate any new memory. It points to the
same memory, same data. If we think of D arrays as something like
this:
struct Array(T) {
size_t len;
T* ptr;
}
Then representation is doing this:
Array original;
Array representation(original.len, original.ptr);
So, yes, the char data will still be shuffled in place. All
you're doing is getting a ubyte view onto it so that it can be
treated as a range.
