Access violation using chain()

[Brad Anderson]

On Thu, Apr 19, 2012 at 7:25 AM, Christophe <travert at phare.normalesup.org>wrote:

> "Brad Anderson" , dans le message (digitalmars.D.learn:34902), a écrit :
> > Perhaps I'm just misunderstanding something about closures but
> > the following code seems to behave oddly:
> >
> >      import std.stdio, std.range, std.algorithm, std.string;
> >
> >      void main()
> >      {
> >          auto lst = ["a", "b"];
> >          auto rng = range_gen(lst);
> >          writeln(rng.take(5));
> >      }
> >      auto range_gen(string[] lst)
> >      {
> >          auto a = sequence!"n+1"().map!(a=>format("%s%d", lst[0],
> > a))();
> >          return chain(lst, a); // access violation
> >          //return a; // works
> >      }
>
> My guess is that chain takes lst by reference, just like the delegates
> for map, wo both are working on the same slice instance. The chain first
> pops elements from lst, and then calls the mapped sequence. At that
> time, lst is empty.
>
> You can just copy lst before you give it to chain (or to map's delegate)
> to solve this bug:
>
> auto range_gen(string[] lst)
> {
>     auto a = sequence!"n+1"().map!(a=>format("%s%d", lst[0], a))();
>      string[] lst2 = lst;
>     return chain(lst2, a); // access violation
> }
>
>
Ah, that would make sense.  I'll test and make sure when I get home. Range
consumption tricks me more often than I wish.  I'll eventually learn to
look out for it more actively.

Regards,
Brad Anderson
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