On Thu, 22 Sep 2011 16:09:29 -0400, Steven Schveighoffer
<[email protected]> wrote:
On Thu, 22 Sep 2011 15:44:21 -0400, Jonathan M Davis
<[email protected]> wrote:
On Thursday, September 22, 2011 23:36:40 Dmitry Olshansky wrote:
On 22.09.2011 22:53, Jesse Phillips wrote:
> The discussion on Reddit brought to my attention that pure functions
can
> return and assign to an immutable.
>
>
http://www.reddit.com/r/programming/comments/knn5p/thoughts_on_immutabil
> ity_in_d/c2lsgek
>
> I am trying to modify the example request to make use of this, but
have
> failed.
>
>
http://www.reddit.com/r/programming/comments/knn5p/thoughts_on_immutabil
> ity_in_d/c2lrfpm
>
> test.d(4): Error: cannot implicitly convert expression
> (makeFromArray([1,2,3])) of type test.List!(int).List to
> immutable(List)
>
> Is this a bug? I can't identify where this issue would lie (works
with
> inheritance and templating).
Maybe:
-------------------------<<<<<<<<<<
List!T makeFromArray(T)(immutable T[] array) pure {
> if (array.length == 0) { return null; }
>
> auto result = new Cons!T(array[0], null);
> auto end = result;
>
> for (int i = 1; i< array.length; ++i) {
>
> end.tail_ = new Cons!T(array[i], null);
> end = end.tail_;
>
> }
>
> return result;
>
> }
If I'm not mistaken only strongly pure functions are working.h
Which would make sense. The only reason that it can implicitly cast to
immutable is because it _knows_ that there are no other mutable
references to
that data, and for it to be able to know that, the function must be
strongly
pure.
Technically, something like this could be cast to immutable:
T[] foo(const(T)[] arg) pure
Since it can be proven that the result is new data.
So it doesn't *need* to be strong-pure.
And actually, just making the argument immutable doesn't make it
strong-pure, the result has to be too.
So I don't think it has to do with strong-purity at all.
-Steve
