in reply to Re: Celsius to Fahrenheit using s///
in thread Celsius to Fahrenheit using s///

Almost as easy to fix it as to find the problem ;->

use strict;
use warnings;
# From http://prometheus.frii.com/%7Egnat/yapc/2000-stages/slide31.htm
+l
# Fixed a typo in the regex (Nathan forgot to escape the forwardslash.
+)
# Fixed number problems - allow +/-, and decimals.  Exponentials still
+ not allowed.

while (<DATA>) {
  chomp;
  my $celsius = my $fahrenheit = $_;
  $fahrenheit =~ s|([+-]?(?:\d*\.)?\d+)C|($1*9/5)+32 . "F"|ge;
  print "$celsius is $fahrenheit\n";
}
#Outputs: 
#12C is 53.6F
#13C is 55.4F
#14C is 57.2F
#15C is 59F
#-40C is -40F
#+32C is 89.6F
#.5C is 32.9F
#15.8C is 60.44F

__DATA__
12C
13C
14C
15C
-40C
+32C
.5C
15.8C

[download]

Replies are listed 'Best First'.
Re^3: Celsius to Fahrenheit using s///
by saskaqueer (Friar) on Feb 08, 2005 at 18:15 UTC

    Using the /e modifier of a regex for such uses is just plain wrong. It looks more like you are obfuscating code than writing it. Much better written in a form such as the following (mine will also do reverse conversions -- F to C): 

    #!/usr/bin/perl -w

use strict;

while (<DATA>) {
    chomp;
    next unless (m!\A([+-]?(?:\d*\.)?\d+)([CF])\z!);

    print "$_ is ", (
        $2 eq 'C' ?
            $1 * 9 / 5 + 32 . 'F' :
            ($1 - 32) * 5 / 9 . 'C'
    ), $/;
}

__END__
12C
13C
14C
15C
-40C
+32C
.5C
15.8C
60.44F
55.4F
-.7F

    [download]
[reply]
[d/l]
      Is this better, or worse? 
      while (<DATA>) {
  chomp;
  my $was = $_;
  my %formulae = (C => sub { ((shift() * 9/5) + 32).'F' },
                  F => sub { ((shift() - 32) * 5/9).'C' });
  s{(.*)([CF])}{${\$formulae{$2}($1)}}; # Update: look, ma! No "e"
  print "$was is $_\n";
}
__DATA__
12C
13C
14C
15C
-40C
+32C
.5C
15.8C
60.44F
55.4F
-.7F

      [download]
      Caution: Contents may have been coded under pressure.
[reply]
[d/l]
      It looks more like you are obfuscating code than writing it.

      Come on, we're converting Celsius to Fahrenheit in a regexp here! :^)

[reply]
        No. You're using the substitution operator. Only the first portion is a regexp.
[reply]

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