I was looking at the code for Scalar::Util, and in specific at the source for the looks_like_number() function. There's something strange in the code that prompts me to ask the question of 'why'.

Here is the source for the looks_like_number() function, pasted from Scalar::Util:

sub looks_like_number {
  local $_ = shift;

  # checks from perlfaq4
  return $] < 5.009002 unless defined;
  return 1 if (/^[+-]?\d+$/); # is a +/- integer
  return 1 if (/^([+-]?)(?=\d|\.\d)\d*(\.\d*)?([Ee]([+-]?\d+))?$/); 
      # a C float
  return 1 if ( $] >= 5.008 and /^(Inf(inity)?|NaN)$/i) 
                or ($] >= 5.006001 and /^Inf$/i );

  0;
}

[download]

The line I'm questioning is: 

  return $] < 5.009002 unless defined;

[download]

As I understand from reading perlvar, $] contains the version + patchlevel / 1000 of the perl interpreter. The question is why is Scalar::Util's looks_like_number() written to return truth if its parameter is undefined and the version is less than 5.009002? This would mean that the looks_like_number() function will errantly report an undefined value as looking like a number if you're using a sufficiently old version of Perl5. Is this really what is intended? Or am I missing something simple?


Dave

In reply to Scalar::Util 'looks_like_number()' code's $] comparison by davido

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