So I stole your test here for a moment...

say join "\n", grep { sprintf ("%03d",$_) + 0 ne ($_/1000)*1000 } 0..9
+9999
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I needed to add a + 0 to be able to handle the trailing zero's. The funny thing is that in your case you would need to do the opposite ( "" . value) in case the value undergoes an arithmetic operation.

This boils down to two alternatives to compare two numeric scalars with a minimal* rounding tolerance

 
 DB<65> $num1 = '0019.90'; $num2 = '019.9'

 DB<66> p $num1+0 eq $num2+0
1

 DB<67> p "$num1" == "$num2"
1

 DB<68>
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Not sure which one is technically better.

I find the second one more readable and easier to explain.

edit

Well there are edge cases when the scalar is not purely numeric... but that's too complicated.

update

FWIW: the eq form can be further simplified with a unary +:

  DB<109> printf "%.20f\n",$_ for $a,$b
2547200.00000000000000000000
2547199.99999999950000000000

  DB<110> p +$a eq +$b                           # unary +
1
                                                                      
+          
  DB<111> p $a == $b                             # doesn't work


  DB<112> p "$a" == "$b"                         # preferred variant
1
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Cheers Rolf
_{(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery}

*) should be noted that this can't help fixing accumulated rounding errors.