On Tuesday, 21 October 2014 at 12:30:30 UTC, anonymous wrote:
On Tuesday, 21 October 2014 at 12:08:35 UTC, Solomon E wrote:
On Tuesday, 21 October 2014 at 08:48:09 UTC, safety0ff wrote:
const int[] a;
int[] b;
static this()
{
b = [1];
a = b;
}
`a` isn't a reference to `b`. `a` is assigned by value and has
its own storage.
`a` is indeed a copy of `b`. But `b` is a pointer+length, and
only those are copied. The array data is not copied. `a` and `b`
refer to the same data afterwards.
[...]
const int[] a;
int[] b;
static this()
{
[...]
a = b;
}
[...]
void main()
{
[...]
b = [8,7];
Here, making `b` point somewhere else (to [8, 7]). If instead
you
change b's elements, you'll see that `a` and `b` refer to the
same data:
b[] = 8; /* Will also change `a`'s data. */
You're right. Thank you, anonymous stranger.
Sorry about that, safety0ff. It looks like you were right and I
was wrong.
`b[0] = 8;` or `b[] = 8;` changes a. Printing the values for &a
and &b shows they're different pointers, but (a is b) returns
true. So I still have more to learn about how it does that.
