Hi,
Today I discovered that floating point errors seem to be much more of a problem than I had anticipated.
Specifically, it seems that if I set up a simple loop to add 0.05 to a variable and then display the result, it quickly goes wrong.
For example, the following code:
my $c = 0;
for ($x = 0; $x <= 1; $x += 0.05){
$c++;
print "[$c] x= $x (".sprintf("%20.40f",$x).") \n";
}
produces this:
[1] x= 0 (0.0000000000000000000000000000000000000000)
[2] x= 0.05 (0.0500000000000000027755575615628913510591)
[3] x= 0.1 (0.1000000000000000055511151231257827021182)
[4] x= 0.15 (0.1500000000000000222044604925031308084726)
[5] x= 0.2 (0.2000000000000000111022302462515654042363)
[6] x= 0.25 (0.2500000000000000000000000000000000000000)
[7] x= 0.3 (0.2999999999999999888977697537484345957637)
[8] x= 0.35 (0.3499999999999999777955395074968691915274)
[9] x= 0.4 (0.3999999999999999666933092612453037872910)
[10] x= 0.45 (0.4499999999999999555910790149937383830547)
[11] x= 0.5 (0.4999999999999999444888487687421729788184)
[12] x= 0.55 (0.5499999999999999333866185224906075745821)
[13] x= 0.6 (0.5999999999999999777955395074968691915274)
[14] x= 0.65 (0.6500000000000000222044604925031308084726)
[15] x= 0.7 (0.7000000000000000666133814775093924254179)
[16] x= 0.75 (0.7500000000000001110223024625156540423632)
[17] x= 0.8 (0.8000000000000001554312234475219156593084)
[18] x= 0.85 (0.8500000000000001998401444325281772762537)
[19] x= 0.9 (0.9000000000000002442490654175344388931990)
[20] x= 0.95 (0.9500000000000002886579864025407005101442)
which is... okayish.. because the errors only show up when I force sprintf to display to 40dp... the problem is that even though I'm only adding 0.05 (and not 0.00000000000000000000000005), the error still creeps into the results pretty quickly.
If I change the loop to run from 0 to 4, which is 100 interations, perl tells me at iteration 74 that 3.65 + 0.05 = 3.69999999999999.
I'm not sure what to think of this.. I wouldn't have expected that adding 0.05 to a number 74 times would start to produce errors as soon as the 1st decimal place.
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