On Thursday, April 18, 2002, at 03:37 , Mark Anderson wrote:
>
>>>> No, you need ++ vs +1. As they say in perl, ++ is magical and will
>>>> do want you want. + 1 will not.
>>>>
>>>> Wags ;) ps -- is not magical in the same sense as ++ either.
>>>
>>> How would you decrement a character then? There surely has to be a
>>> way?
>>
>> perldoc -f ord
>> perldoc -f chr
>>
>> my $len = $#array ;
>> my $x = 0;
>> $array[$x++] = chr( ord($array[$x]) - 1 ) while( $x <= $len );
>
> This decrements ever character in an array. The ++ is magical in that "a"
> increments to "b" and "z" increments to "aa". (ultimately leading to a
> cute
> little japh)
>
> Your solution takes "abcde" and returns "`abcd" which is not the opposite
> of
> the ++ operator.
actually no.... the code and run of the code is shown in details below....
one could rewrite
$array[$x++] = chr( ord($array[$x]) - 1 ) while( $x <= $len );
as
while( $x <= $len ) {
my $letter = $array[$x]; # assign nth letter
my $val = ord($letter); # get it's ordinal value
$val--; # decrement the ordinal
$array[$x] = chr($val); # insert the Char back in
x++; # increment on down
the way.
}
and it may be a bit more obvious that we are walking ourselves
through an 'array of letters' - since the question had been posited
in regards to decrementing letters....
sorry if the flatlining of it was not as obvious.
[..]
#!/usr/bin/perl -w
use strict;
my @array = qw/ a b c d e f g h i j k l m n o p /;
print "Letters are now @array \n";
my $len = $#array ;
my $x = 0;
$array[$x++] = chr( ord($array[$x]) - 1 ) while( $x <= $len );
print "Letters are now @array \n";
$array[$x]++;
print "Letters are now @array \n";
[jeeves:/tmp/drieux/perl] drieux% perl !$
perl pr*
Letters are now a b c d e f g h i j k l m n o p
Letters are now ` a b c d e f g h i j k l m n o
Letters are now ` a b c d e f g h i j k l m n o 1
[jeeves:/tmp/drieux/perl] drieux%
ciao
drieux
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