$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]

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

I am not so sure about that.

print "1989.9999999999998" + 0 ; # 1990

The 'target form' is decimal string.

From perlnumber Target form: If the source number can be represented in the target form, that representation is used.

Also: When a numeric value is passed as an argument to such an operator, it will be converted to the format understood by the operator.. My assumption is that this counts for  + - / * etc applied to the decimal string and eventually also for eq.

The print is showing the string representation only, where rounding errors are smoothed away in a DWIM way.

Try printf "%.20f",$var to see the rounding errors pertaining after +0.

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