Why TypeTuple can be assigned to a variable

[Zhenya]

On Wednesday, 12 June 2013 at 08:14:06 UTC, Simen Kjaeraas wrote:
> On Wed, 12 Jun 2013 10:01:59 +0200, Zhenya <zheny at list.ru> 
> wrote:
>
>> Hi!
>>
>> I was just surprised when realized, that this code compiles 
>> and runs:
>>
>> import std.typetuple;
>> import std.stdio;
>>
>> void main()
>> {
>>  auto foo = TypeTuple!("foo","bar");
>> writeln(typeid(typeof(foo)));
>>  writeln(foo);
>> }
>>
>> If I were compiler expert,I'd say that it's a bug.But I am not)
>> So, can anybody explain why it's work?
>
> It is the equivalent of:
>
>   TypeTuple!(string, string) foo;
>   foo[0] = "foo";
>   foo[1] = "bar";
>
> The ability to have a tupetuple as a variable is very useful - 
> if that
> had not been possible one would need something akin to this:
>
> struct Tuple(T...) {
>     T[0] head;
>     static if (T.length) {
>         Tuple!(T[1..$]) tail;
>     }
> }
>
> And to access the third element of a tuple one'd need to write:
>
>     myTuple.tail.tail.head = "foo";
>
> Clearly this is suboptimal, so D has better ways of doing such 
> things.


OK,you say that TypeTuple!("foo","bar") is a cool value of type
TypeTuple!(string,string),right?

Then could you tell me what type has [TypeTuple!("foo","bar")]?

Currently it is evaluated to string[].And I think if TypeTuple

had value semantic, this type would be TypeTuple!(...)[].

This behaviour confuses me a bit.

And I just don't understand why do we need TypeTuple's value
semantic
to implement std.Tuple,because AFAIK it use variadic template
parameter pack
   != TypeTuple.
Sorry for my english.