On Saturday, 23 February 2013 at 20:28:43 UTC, Steven Schveighoffer wrote:
On Sat, 23 Feb 2013 10:34:18 -0500, deadalnix <[email protected]> wrote:

On Saturday, 23 February 2013 at 15:16:15 UTC, Steven Schveighoffer wrote:
This actually is impossible to do with inout, unless your ranges are pointers or arrays (believe me, I tried with dcollections). But that is not inout's fault, it's because we have no way to specify tail-const arbitrary types.


I have read what come after, but clearly shouldn't have. This is the interesting thing to discuss. Can you expand on that ? Some code sample ?

OK, here is a simple abstract example (the implementation details are not important, just the interface).

class Container(T)
{
   ...
   struct range
   {
      ...
      @property inout(T) front() inout {...}
      void popFront() {...}
      bool empty() const { ...}
   }

   inout(range) opSlice() inout {...}
}

Nice, right? But it doesn't work, for one simple reason: popFront.

const Container!int cont = new Container!int(...);

auto r = cont[];

foreach(const x; r)
{
   writeln(x);
}

doesn't work! typeof(r) is const(Container!(int).range), and since popFront is not const (and cannot be), you cannot iterate!

What you need is a SEPARATE "const_range" type (this is similar to C++). Then opSlice() const will return const_range.

And const range and range are not related, they are separate types. So there is no way to have inout choose the right one!

HOWEVER, if your range is a slice it works beautifully, because you can do a tail-inout slice inout(T)[].

Not possible with custom types.

I am working on a plan on how to fix this, because it really annoys me for my project so much to duplicate code that I just don't even support const ranges or immutable ranges! I do support immutable and const cursors, since those point at one element.

But I have not gotten around to pitching it because I need to come up with a really good solution :) Walter is so sour on any tail-const solution from past attempts that it has to be bullet-proof and easy.

-Steve

Steven would you mind to create a topic on this subject precisely ?

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