Re: First JAPH - Spell perl in two hundred and eighty five thousand and seventy four easy steps

[pokemonk]

I find this really amazing, but i can't figure out why it happens. can someone please explain?

pokmon

[blakem]

From perlop:
The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern /^[a-zA-Z]*[0-9]*$/, the increment is done as a string, preserving each character within its range, with carry:
print ++($foo = '99'); # prints '100'
print ++($foo = 'a0'); # prints 'a1'
print ++($foo = 'Az'); # prints 'Ba'
print ++($foo = 'zz'); # prints 'aaa'


The auto-decrement operator is not magical.

-Blake

[Anonymous Monk]

#!/usr/bin/perl
use strict;
my $theword = @ARGV[0];
my $A='a';
FINDIT: for (0..99999999){
$A++;
if ($A eq "$theword")
    {print "$_\n";
    last FINDIT;
    }
}

[holonaut]

Wow. That can take a really long time to count. This little bit o' code is much faster.

 #!/usr/bin/perl -w

# Convert pseudo-base26 number (digits: a-z) to decimal
# really intended to find the ending number needed in
# the for loop for any given string to be used

my $num = 0;
my $char = "a";
my $answer = 0;

until ($char eq "aa") {
    $equiv{$char}=$num;
    $num++; $char++;
}

print "Enter your lowercase string: ";
chomp($string = <STDIN>);
@strArray=reverse(split(//,$string));

for ($i = 0; $i < @strArray; $i++) {
    $answer += ($equiv{$strArray[($i)]} * (25 ** $i));
}

print --$answer;

[admiraln]

I tried the code and it did not work. After a little experimenting following code did work. I added the orignal one liner to the generator to validate it. The mistakes that I saw were two fold. First base should have be 26 not 25. Second the magical text increment behavior is not equivalent to incrementing numbers.

9 goes to 10
z goes to aa  not a0 

as a result the equivs have to run a-z === 1-26 not 0-25 and the answer offset is 2 not 1.

#!/usr/bin/perl -w

# Convert pseudo-base26 number (digits: a-z) to decimal
# really intended to find the ending number needed in
# the for loop for any given string to be used

my $num = 1;        #  was  $num = 1
my $char = "a";
my $answer = 0;

until ($char eq "aa") {
    $equiv{$char}=$num;
    $num++; $char++;
}

print "Enter your lowercase string: ";
chomp($string = <STDIN>);
@strArray=reverse(split(//,$string));

for ($i = 0; $i < @strArray; $i++) {
    $answer += ($equiv{$strArray[($i)]} * (26 ** $i)); 
                                     # was 25 ** $i   
}
$offset = $answer - 2;
print $offset,"\n";

my $A="a";for(0..$offset){$A++;}print"$A";

[kiat]

hi pokemonk;

In order to get an idea of how the word 'perl' gets printed, you may want to modify the code as follows:



#!c:\perl\perl.exe -w
use strict;
my $A="a";for(0..285074){$A++; print"$A";}

That is, you place the 'print "$A"' statement inside the loop. To see what what happens when, say, 1000 characters are printed, you stop the loop by pressing Ctr 'C' (if you're using Windows).

kiat