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Re: Re: Hash slices ?

by snax (Friar)
on Dec 01, 2000 at 17:45 UTC ( #44340=note: print w/replies, xml ) Need Help??

in reply to Re: Hash slices ?
in thread Hash slices ?

Ayup -- it's the last key that matters, just as if the hash slice were being treated as a list -- evaluate a list in a scalar context and it returns the last value.

There's nothing in the docs that I could find that says that a hash slice would be interpreted as a list rather than a proper array, but it does apear to be the case.

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Re: Re: Re: Hash slices ?
by davorg (Chancellor) on Dec 01, 2000 at 17:53 UTC

    Any one further investigation, it seems that my original solution suffered from a fundamental problem.

    my %hash = (one => 1, two => 2, three => 3); my @keys = qw(four five six); my @slice = @hash{@keys};

    @slice is now contains four undef elements, where I thought it would be an empty list (don't know why I thought that - it was very early in the morning.

    If the hash slice was interpreted as an array, it would therefore always be true and therefore doesn't solve ChOas' original problem. That would be better addressed using something like:

    if (grep { exists %hash{$_} } @keys) { print "yep\n"; } else { print "nope\n"; }

    "Perl makes the fun jobs fun
    and the boring jobs bearable" - me

      perl -w -e '%x=(a=>7,b=>8,c=>9);@s=(qq(a),qq(b));$y=@x{@s};print $y;'
      returns 8 -- so it is pretty clearly the case that a hash slice is properly a list.

      I was thinking about doing it like this to avoid the grep:
      sub Match{foreach (@keys) {return 1 if exists $hash{$_};}return 0} if ( Match ) { print "yep\n"; } else { print "nope\n"; };
      But I was thinking about that already before I thought of
      the slice ;))
Re: Re: Re: Hash slices ?
by merlyn (Sage) on Dec 01, 2000 at 20:14 UTC
    Well, we always go around about this, because the perldata does clearly say:
    @days # ($days[0], $days[1],... $days[n]) @days[3,4,5] # same as @days[3..5] @days{'a','c'} # same as ($days{'a'},$days{'c'})
    See the "same as"? That's the operative words. To make it the same, it has to return the last element of the list in a scalar context. So yes, the behavior is documented, and derivable from the docs.

    -- Randal L. Schwartz, Perl hacker

      Poppycock! That is such an outrageous statement I don't really know where to start.

      If we are to take "same as" in some comment in some sample code and extrapolate fundamental behavior from it... So since "same as" has such profound meaning, then it must also be true that these two are "the same":

      @days{@x,@y} ($days{@x},$days{@y})
      or even these:
      @days{foo(),bar()} ($days{foo()},$days{bar()})
      when in fact, they aren't.

      The fact is that a hash slice in a scalar context returns the last item of the hash slice because there is C code that specifically does that. What you quoted doesn't qualify as documentation of that fact. If that is the only documentation, then it is undocumented.

      In fact, Perl actually goes though and builds the entire list result of the hash slice and then moves the last item to the front:

      if (GIMME != G_ARRAY) { MARK = ORIGMARK; *++MARK = *SP; SP = MARK; }
      unlike many "list-returning operations" that know to optimize their behavior in the face of scalar context. And all of this has very little to do with the comma operator which you claim it is "documented" as being "same as".

              - tye (but my friends call me "Tye")
        There is a significant and important difference between how you are supposed to think about things and what goes on under the hood.

        merlyn is absolutely right about what concepts Perl's behaviour is intended to give. You are absolutely right about the implementation. It is both possible and likely that a future implementation will render your observation incorrect. It is much less likely that the concept both merlyn and perldata are trying to convey will be invalidated at a future point.

        Therefore I like merlyn's answer and do not consider it, implementation trivia notwithstanding, poppycock.

      First, thanks for your comment.

      Now, on to what I originally wrote, which may sound snippy -- it's not meant to be.

      Right. Well. That's a bit cryptic, though, no?

      Further, 2nd Ed of Programming Perl provides that the first line is "Same as" as well. I'm not here to nitpick about right or wrong, though. Perhaps a line in perldata someplace that says "hash slices are lists" would clear things up.

      Honestly, I find this behavior a bit odd. With a list, evaluated in a scalar context so that it returns its last value, you could at least have things going on prior to the last value that are meaningful in determining the last value (much like constructs in C with the comma operator).

      By the time you shove stuff into an array, or hash, and then pull out a slice, it makes more sense to me for this to be more "array like" than "list like"; the values before the last one are then in the dreaded void context if this object is then evaluated in a scalar context.

      That is definitely enough musing from me about what is, to me, a very dusty little corner of perl.

      True, it is derivable from the docs. Having just finished a conversation with a co-worker about another nearly undocumented feature of Perl, I find myself just a bit frustrated with these "derivable" documentation tidbits.

      The difference between arrays and lists can be pretty esoteric. In this case, it takes a pre-existing understanding of the list vs. array issue to decode the docs.

      As a frequent Perl mentor, I often answer this type of question. Where can I find a good reference that will point out issues like this? Effective Perl Programming is a fantastic "up-and-coming-Perl-coder" book. It doesn't seem to address this lists vs. array issue, however. Are there any other good intermediate-level books which cover this kind of "gotcha?"

      Brainbench 'Most Valuable Professional' for Perl

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