On Saturday, July 28, 2012 02:49:16 Chad J wrote:
> Is there some way to do something similar to this right now?
>
> void main()
> {
> // Differing levels of type-inference:
> int[] r1 = [1,2,3]; // No type-inference.
That works just fine.
> Range!(int) r2 = [1,2,3]; // Only range kind inferred.
What do you mean by range kind? If you declare Range!int, you gave it its type
already. It's whatever Range!int is. You can't change it. There's nothing to
infer. Either Range!int has a constructor which takes an int[] or the
initialization won't work. Range!int is the same either way.
> Range r3 = [1,2,3]; // Element type inferred.
Again, the type of the variable must already be a full type, or you can't
declare it. So, there's nothing to infer. Either Range!int has a constructor
which takes an int[] or the initialization won't work. Range is the same
either way.
> auto r4 = [1,2,3]; // Full type-inference.
This works;
> AFAIK it isn't: the type system is pretty much all-or-nothing about
> this. Please show me that I'm wrong.
The _only_ time that the type on the left-hand side of an assignment
expression depends on the type of the right-hand side is if the type is being
explicitly inferred by using auto, const, immutable, or enum as the type by
themselves. Types are never magically altered by what you assign to them.
If you want to, you can create a templated function which picks the type to
return. e.g.
Type var = makeType([1, 2, 3]);
or
auto var = makeType([1, 2, 3]);
but it's the function which determines what the type is.
- Jonathan M Davis
