I'm having trouble understanding this code in std.range.chainImpl:

        @property ChainImpl save()
        {
            auto result = ChainImpl();
            foreach (i, Unused; R)
            {
                result._input[i] = _input[i].save;
            }
            return result;
        }

Specifically, line 1559 which is:
auto result = ChainImpl();

Let me see if I got my hunch right:

Is this simply instantiating another implementation of ChainImpl with the same arguments? e.g. inside an already instantiated template with ChainImpl(Range1, Range2), would ChainImpl() simply instantiate another ChainImpl template with the same compile-time arguments? Of course, it wouldn't copy any runtime values since this is all done at compile-time and they don't exist yet.

Btw, I'm having a kick-ass time understanding Phobos as of lately. It only took me a couple of months but I am slowly grokking templates, even though I've *never* used templates before I tried D.

On Wed, 29 Dec 2010 15:26:25 -0500, Andrej Mitrovic <none@none.none> wrote:

> I'm having trouble understanding this code in std.range.chainImpl:
>
>         @property ChainImpl save()
>         {
>             auto result = ChainImpl();
>             foreach (i, Unused; R)
>             {
>                 result._input[i] = _input[i].save;
>             }
>             return result;
>         }
>
> Specifically, line 1559 which is:
> auto result = ChainImpl();
>
> Let me see if I got my hunch right:
>
> Is this simply instantiating another implementation of ChainImpl with the same arguments? e.g. inside an already instantiated template with ChainImpl(Range1, Range2), would ChainImpl() simply instantiate another ChainImpl template with the same compile-time arguments? Of course, it wouldn't copy any runtime values since this is all done at compile-time and they don't exist yet.

Yes.  Inside a template, the name of the template is synonymous with the template instance being instantiated.

So for example:

struct S(T)
{
   void foo(){ S s; // this is of type S!T
   }
}

-Steve

On Wed, 29 Dec 2010 15:35:31 -0500
"Steven Schveighoffer" <schveiguy@yahoo.com> wrote:

> Yes.  Inside a template, the name of the template is synonymous with the template instance being instantiated.
> 
> So for example:
> 
> struct S(T)
> {
>     void foo(){ S s; // this is of type S!T
>     }
> }

Good to know ;-) I write specialised types everywhere in template, like

struct S(T) {
    S!T aNewSBangT (S!T anotherSBangT) {...}
}

Denis
-- -- -- -- -- -- --
vit esse estrany ☣

spir.wikidot.com

One more question. What does this do?

static RvalueElementType fixRef(RvalueElementType val)
{
    return val;
}

This looks like a workaround for something but I can't figure out why the function is static and why does it just return the value passed in?

On 12/29/10, spir <denis.spir@gmail.com> wrote:
> On Wed, 29 Dec 2010 15:35:31 -0500
> "Steven Schveighoffer" <schveiguy@yahoo.com> wrote:
>
>> Yes.  Inside a template, the name of the template is synonymous with the template instance being instantiated.
>>
>> So for example:
>>
>> struct S(T)
>> {
>>     void foo(){ S s; // this is of type S!T
>>     }
>> }
>
> Good to know ;-) I write specialised types everywhere in template, like
>
> struct S(T) {
>     S!T aNewSBangT (S!T anotherSBangT) {...}
> }
>
> Denis
> -- -- -- -- -- -- --
> vit esse estrany ☣
>
> spir.wikidot.com
>
>

On Wed, 29 Dec 2010 17:35:05 -0500, Andrej Mitrovic <andrej.mitrovich@gmail.com> wrote:

> One more question. What does this do?
>
> static RvalueElementType fixRef(RvalueElementType val)
> {
>     return val;
> }
>
> This looks like a workaround for something but I can't figure out why
> the function is static and why does it just return the value passed
> in?

A static function inside a struct or class does not have a 'this' reference.  It's like a free function (one outside a class or struct definition) but is within the namespace of the struct or class.

Why it is there?  I think because a chain range's components may all return by ref, or at least one may return by value.  If at least one returns by value, it logically follows that the chain range itself must return by value.  If all return by ref, it's ok for the chain range to return by ref.

So why does the fixRef exist?  Probably because the compile-time checks used to determine whether you can return by ref or return by value are verbose, and fixRef is used in many places.  I'm only speculating, I didn't write it.  With proper compiler inlining, the fixRef function gets optimized out, but I think ref functions do not get inlined at the moment.

-Steve