[erlang-questions] Re: List comprehension questions
James Churchman
jameschurchman@REDACTED
Fri Apr 1 19:24:22 CEST 2011
So if (from the docs) list comprehension become :
'lc^0'([E|Tail], Expr) -> [Expr(E)|'lc^0'(Tail, Expr)];
'lc^0'([], _Expr) -> [].
then, what happens when Expr contains references to other variables in the original scope? (eg not E ) it says Expr used to be a Fun, but now it's not, so in this case is it a Fun, or does it increase the arty of the generated lc^0 function for each of the extra referenced vars?
Also keen to know the body recursive "cost", and why it avoids using list reverse in the case when a list is required to be returned :-)
james
On 1 Apr 2011, at 12:15, Olivier BOUDEVILLE wrote:
>
> Hello Richard,
>
> First of all, sorry, I had indeed missed your answer. Actually I did not see it neither at work nor at home, and I wonder whether there is not a mailing-list problem as it does not seem to appear either in gmane or in the thread as shown (at least at the time of this writing) by google groups (http://groups.google.com/group/erlang-programming/browse_thread/thread/8d339e24ed090311).
>
> Back on topic: as I understand now (I am not too familiar with Core Erlang), list comprehensions generate local functions that are body-recursive, i.e. involve actually a direct recursion, not using an accumulator.
>
> I just happened to find http://www.erlang.org/doc/efficiency_guide/listHandling.html#id61285 which gives a similar explanation (maybe a link from the reference manual to the efficiency guide could be made, to associate these two list comprehension topics).
>
> I was wondering if this body-recursiveness did not imply that a list comprehension operating on a longer list, whose result would be kept, could eat a lot of stack space/take longer to execute than a tail-recursive version?
>
> Apparently according to http://www.erlang.org/doc/efficiency_guide/myths.html#id58884 this will probably not be the case, so using list comprehension does not involve much trade-off.
>
> Thanks again,
> Best regards,
>
> Olivier Boudeville.
> ---------------------------
> Olivier Boudeville
>
> EDF R&D : 1, avenue du Général de Gaulle, 92140 Clamart, France
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>
> carlsson.richard@REDACTED
> 01/04/2011 10:46
>
> A
> erlang-questions@REDACTED, olivier.boudeville@REDACTED
> cc
> Objet
> Re: [erlang-questions] Re: List comprehension questions
>
>
>
>
>
> On 04/01/2011 10:00 AM, Olivier BOUDEVILLE wrote:
> >
> > Hello Robert,
> >
> > Thanks for your explanation. I guess that the reversal (to restore the
> > original list order after a first "if selected by filter then apply"
> > traversal) is done then in the list:map/2 counterpart!
> >
> > My concern was that, for long lists which do not happen to have to
> > respect any particular order in their elements, two traversals would be
> > done with list comprehensions whereas just one could be sufficient.
> > Well, a minor concern as for such cases we can write our own functions
> > to do so in one pass.
>
> Perhaps you didn't see my reply to your original mail (repeated below).
> The point that both Robert and I are trying to make is that there is no
> reversal involved in the code, and it only does one traversal, so you
> don't need to worry about this.
>
> /Richard
>
>
> On 03/30/2011 01:29 PM, Olivier BOUDEVILLE wrote:
> > Reading
> > http://www.erlang.org/doc/reference_manual/expressions.html#id77052 it
> > is not obvious that the order induced by a single generator is
> > necessarily preserved by the comprehension, as nevertheless suggested by
> > all experiments like:
> >
> > 1> [ X || X <- [1,2,3] ].
> > [1,2,3]
> >
> > I suppose this order preservation is true. Then, I imagine that,
> > internally, list comprehensions are translated to code very similar to
> > the one that would be produced if the developer had written an
> > appropriate accumulator-based tail-recursive function?
> >
> > If so, does it imply that using a list comprehension on a semantically
> > unordered set of elements will involve a useless reversal (like
> > lists:reverse/1) of the resulting list?
>
> No, the translation does a straightforward recursion over the list
> without accumulator, automatically preserving order. To see what
> happens, take for example the following module:
>
> -module(lc).
> -export([f/1]).
> f(Input) ->
> [N + 1 || N <- Input].
>
> and compile it to Core Erlang with c(lc,[to_core]) or erlc +to_core
> lc.erl. The resulting file lc.core contains the following code:
>
> 'f'/1 =
> fun (_cor0) ->
> letrec
> 'lc$^0'/1 =
> fun (_cor3) ->
> case _cor3 of
> <[N|_cor2]> when 'true' ->
> let <_cor4> =
> call 'erlang':'+'
> (N, 1)
> in let <_cor5> =
> apply 'lc$^0'/1
> (_cor2)
> in ( [_cor4|_cor5]
> -| ['compiler_generated'] )
> <[]> when 'true' ->
> []
> ( <_cor3> when 'true' ->
> primop 'match_fail'
> ({( 'function_clause'
> -| [{'name',{'lc$^0',1}}] ),_cor3})
> -| ['compiler_generated'] )
> end
> in apply 'lc$^0'/1
> (_cor0)
>
> (The annotations ( Expr -| 'compiler_generated') on some subexpressions
> are mostly needed to suppress certain compilation warnings that might
> otherwise be triggered, such as for unused results of expressions.)
>
> /Richard
>
>
>
>
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