Re: Big problem : find a number of day.
- From: anno4000@xxxxxxxxxxxxxxxxxxxxxxx (Anno Siegel)
- Date: 4 Nov 2005 10:42:42 GMT
Alextophi <ext.astek.brakha@xxxxxxx> wrote in comp.lang.perl.misc:
> *-* I must find for a number '$day' the day or the days of the table
> %day_array:
>
> my $day = 4;
> my %day_array = ("1" => "0", # Dimanche
> "2" => "1", # Lundi
> "4" => "2", # Mardi
> "8" => "3", # Mercredi
> "16" => "4", # Jeudi
> "32" => "5", # Vendredi
> "64" => "6" # Samedi
> );
>
> my $New_day = $day_array{$day}; # result : $New_day = 2
>
>
> *-* problem: how to make if $day = 5 or 11 or another ?
>
> 5 = 4+1 # result 2 et 0
> 11 = 8+2+1 # result 3 et 1 et 0
>
>
> it is a big problem!
There are straightforward bit-shifting solutions to that. Here is one
that involves a little trick:
my $day_mask = 11;
while ( $day_mask ) {
my $next = $day_mask & ( $day_mask - 1);
print "$day_array{ $next ^ $day_mask}\n";
$day_mask = $next;
}
The expression "$day_mask & ( $day_mask - 1)" deletes (sets to zero)
the least significant bit of $day_mask, leaving the other bits unchanged.
So the XOR of both is the least significant bit in isolation, which is
the index into %day_array.
%day_array isn't even strictly necessary, it is nothing but a coarse
table of logarithms (of base 2). So instead of
print "$day_array{ $next ^ $day_mask}\n";
you could also say
print log( $next ^ $day_mask)/log( 2), "\n";
Rounding the quotient would make it more robust.
Anno
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