Hello Monks!

As we all know, there is no real "number" or "string" data type in Perl, and Perl usually does The Right Thing depending on the context: 

  $x=15;
  $x.=4;
  $x+=2; # ==> 156

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I am now looking for an elegant way of exploiting this in my program, in the following way:

I would like to compare two values for equality so that, iff Perl would agree to see them as a number, numerical equality should be used (i.e. '0' and '000' should be regarded equal), but if at least one of them is not a number, string equality should be used.

The only solution I came up with is very ugly and goes roughly like this: 

$temp1=eval { 0+$value1 };
$temp1=$value1 if $@; Perl says it is not a number
$temp2=eval { 0+$value2 };
$temp2=$value2;
# By virtue of eval, we can now use string equality
# even if the values were numbers before.
if ($temp1 eq $temp2) { ... }

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Needless to say that this is extremely awful. Of course one might argue that what I want to achieve is a bit weird too, but still I would like to know whether there is an easy way to do it. Basically, I'm looking for an operation which would leave a string unchanged, but normalizes a number so that for example '001' would turn into '1' (this would be done by my addition of zero for instance).

-- 
Ronald Fischer <ynnor@mm.st>

In reply to Exploiting Perls idea of what is a number by rovf

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