in reply to storage of numbers

use Scalar::Util qw( looks_like_number );

while( <> ) {
  next unless looks_like_number($_);
  chomp;
  $min = ! defined( $min ) || $min > $_ ? $_ : $min;
  $max = ! defined( $max ) || $max < $_ ? $_ : $max;
}
print "\nminimum number: $min\n";
print "maximum number: $max\n";

[download]

Dave

Replies are listed 'Best First'.
Re^2: storage of numbers
by vsespb (Chaplain) on Oct 16, 2013 at 17:57 UTC
    Do you know that 
    perl -MScalar::Util=looks_like_number -e 'print looks_like_number "Inf
+"'

    [download]
    prints true?
[reply]
[d/l]

      That's not what happens under Win7, perl 5.16, and the latest AS 5.16-repository version of Scalar::Util (with quotes adjusted for MS pleasure):

      C:\>perl -MScalar::Util=looks_like_number -e "print looks_like_number 
+'Inf'"
20

      [download]

      Oh. OK, if you say so. But why did the output ne 'True' (unless 'Inf/INF/inf' is treated as NAN?
      Just then, though, a lightbulb flashed on -- AhHA,   /me said to myself!     RTFM!

      looks_like_number EXPR
Returns true if perl thinks EXPR is a number. See "looks_like_number" 
+in perlapi.

      [download]

      Unfortunately, the link, <a href="/perldoc?perlapi#looks_like_number" class="podlinkpod">&#34;looks_like_number&#34; in perlapi</a> (to which 'See' directs us) is broken.

      So forging blindly onward with what seemed, OTTOMH, some semi-plausible alternatives:

      C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber 'Inf'"
20
C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber 'foo'"
0
C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber 'inf'"
20
C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber '7'"
1
C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber '7874321907'"
2
C:\>perl -e "use Scalar::Util(looks_like_number); print looks_like_num
+ber(7874321907)"
8704

      [download]

      It all leaves me deep in 'WTF' territory and slipping into the quicksand.

      So prithee, Wiser heads: Prevail! and halp, hal blubh blub o o ....

      Update: added first 1.5 sentences after the first blockquoted code.

[reply]
[d/l]
[select]
        When I told "prints true? " I meant something which is TRUE. Yes, it was 20. Sorry for confusion.
[reply]
        Well, anyway, the point was you don't always really want accept "Inf" as number. Sometimes you just want "normal" numbers.
[reply]

      Do you know that...

      Yes, it's documented in perlapi. I didn't assume that would be a concern for the OP.

      The old pure-Perl version of Scalar::Util has a pure-Perl version of looks_like_number, which includes the following code:

      return 1
  if ( $] >= 5.008    and /^(Inf(inity)?|NaN)$/i ) 
  or ( $] >= 5.006001 and /^Inf$/i               );

      [download]

      So if we wanted to eliminate that behavior, we could either copy/paste and modify the entire pure-Perl version from an old version of Scalar::Util (but it has issues with '0 but true'), or use the modern version of Scalar::Util::looks_like_number like this:

      sub looks_like_finite_number {
  my $candidate = shift;
  return 0 if ! looks_like_number( $candidate );
  return 0
    if ( $] >= 5.008    and $candidate =~ /^(Inf(inity)?|NaN)$/i) 
    or ( $] >= 5.006001 and $candidate =~ /^Inf$/i );
  return 1;
}

      [download]

      Or, in perlfaq4 there is a series of regexes given for determining if one is looking at a number under the section, "How do I determine whether a scalar is a number/whole/integer/float? It's not pretty, and that FAQ answer has even changed over the years; it used to use a bunch of "if" statements. Now it uses given/when. *sigh*


      Dave

[reply]
[d/l]
[select]

In Section Seekers of Perl Wisdom