> Because $num2 is not a floating point

$num2 is a float with rounding error after already before the multiplication.

Just the string representation will ignore the rounding error hence eq will work here.

I that's reliable for all cases? I don't dare saying.

 DB<4> $num2 ="19.90"

 DB<5> printf "%.20f", $num2*100
1989.99999999999980000000
 DB<6> p 1990 == $num2 *100

 DB<7> p 1990 eq $num2 *100
1
  DB<8> say ">". $num2 *100 ."<"
>1990<

 DB<9>

[download]

Cheers Rolf
(addicted to the Perl Programming Language :)
Wikisyntax for the Monastery

In reply to Re^2: number comparison with a twist by LanX
in thread number comparison with a twist by anotherguest

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