use Perl

• Time zone problem? (Score:2)

  by vsergu (505)

  The title I'm getting is "Sun 22 May 2005 08:57:22 PM EDT (Mon, 23 May 2005 08:57:22 GMT)". The difference between EDT and GMT is 4 hours, not 12. But perhaps it's some other EDT you're referring to. (Plus it's confusing to use AM/PM for the EDT times and 24-hour time (apparently) for the GMT.)
  • Re:Time zone problem? (Score:2)

    by TorgoX (1933)

    That's very strange. All the time math is done in your browser, which should ideally know its correct TZ offset. But I'll poke around a bit and see if I'm parsing the date wrong or something. What browser are you using?
    • Re:Time zone problem? (Score:2)

      by vsergu (505)

      Firefox on Linux. That was at work. I just tried it at home, also Firefox on Linux, and it's the same.
      • Re:Time zone problem? (Score:2)

        by vsergu (505)

        Argh! If the string begins with '0', parseInt() assumes it's octal, and '08' isn't a valid octal number and thus returns 0. Maybe subtracting 0 (as opposed to adding 0, which will just give string concatentaion) is better than using parseInt() when you're converting strings to numbers.
        • Re:Time zone problem? (Score:2)

          by TorgoX (1933)

          D'oh, and just days after I wrote this!!! [perl.org]

          It seems that if I pass an explicit ",10" to parseInt(), it doesn't do that crazy octalification thing. There, fixedified [interglacial.com]!
• not in HOP (Score:1)

  by brev (1827)

  I haven't even finished HOP yet, but I just digested a large Javascript tome [oreilly.com] for work reasons. JS has a few higher-ordery features that Perl doesn't.

  An interesting twist on eval, Function() creates a function reference out of an array of strings.

  And this JS 1.5 syntax, which gives an anonymous closure a temporary name so it can refer to itself.

  var f = function fact(x) { if (x <=1) return 1; else return x*fact(x-1); }

  Despite appearances, there is no fact(). That was just a name for use duri
  • Re:not in HOP (Score:2)

    by TorgoX (1933)

    One is tempted to do this:

    my $f = sub { ($_[0] <=1) ? 1 : ( $_[0] * $f->($_[0] - 1)) };
    print $f->(4);

    But that doesn't work, because sub is defined outside the scope of $f. So you've got to do this:

    my $f;
    $f = sub { ($_[0] <=1) ? 1 : ( $_[0] * $f->($_[0] - 1)) };
    print $f->(4);

    That works!

    (GC note: now $f and that sub constitute a circular data structure, so you need to do $f='whatever'; (or just undef($f)) to break the circularity.)
    • Re:not in HOP (Score:1)

      by brev (1827)

      Huh. Now I wonder why what I was trying didn't work. That seems relatively straightforward, with the slight inconvenience of defining the variable first. Callee was another weirdness I noticed too. Thanks.
  • And with arguments.callee (Score:2)

    by TorgoX (1933)

    Also:

    var f = function (x) { if (x <=1) return 1; else return x * arguments.callee(x-1); };