Type value

[Zhenya]

On Tuesday, 11 December 2012 at 07:50:10 UTC, js.mdnq wrote:
> On Tuesday, 11 December 2012 at 07:15:38 UTC, Zhenya wrote:
>> I'm sorry for my english.
>> I know that D != Python and my example don't need
>> any RTTI.
>>
>> D has alias this feature.My example shows that we can use 
>> alias this with types.
>> And I just want use this:
>>
>> struct Type(T)
>> {
>>   alias T m_type;
>>   alias m_type this;
>> }
>>
>> int main()
>> {
>>   Type!int Int;
>>   Int i;//should compile by definition of alias this
>> }
>
> My I ask, why not just use int directly?
>
> I'm not proficient enough with D yet but it seems to me that 
> the example does not make sense.
>
> alias T m_type;
> alias m_type this;
>
> should be equivalent to
>
> alias T this;
>
> which, in some sense makes struct Type(T) equal to T... which 
> is what you are thinking, I think.
>
> But! I also believe that structs are meant as value types and 
> are addressable.
>
> So when you do
>
> Int i; you are essentially doing something like
>
> int x;
> x i;
>
> which makes no sense.
>
> If you do this instead,
>
> 	alias Type!int Int;
> 	Int i;
>
> then it will compile, of course.
>
> But your way, when you do Int x; you are specifically trying to 
> create a instance of an object. Yet you are not wanting it to 
> be an object and I'm not sure what you are wanting x to be. 
> (another type? a type of a type?)

In my example Int is value of Type!int.
When I write
Int i;
If Int had not alias this it really would not have sense,because 
type expected
instead of 'Int'.But Int have alias this 'int',and can be 
replaced by it.
My reason is simple:
Operators are functions.But if I want declare some struct like 
std.TypeTuple
I will need opSlice that return type.
It is impossible.
But if my example compiled,I could return value,that can be used 
as type.
It's something like Andrei's Type2Type.