number accuracy

[kencavagnolo]

I was digging through some code this morning after encountering a
program error and I came across the following...

For this bit of code:
#-------------------------------------------------------#
$eta = 0.5;
$etacrit = 0.97;
while ($eta <= $etacrit) {
($eta < 0.80) ? ($etastep = 0.1) :
($eta < 0.95) ? ($etastep = 0.05) :
($etastep = 0.01);
$eta += $etastep;
}
#-------------------------------------------------------#

Which *should* output:
0.6
0.7
0.8
0.85
0.9
0.95
0.96
0.97

It instead was outputing:
0.6
0.7
0.8
0.9
0.95
0.96
0.97

So I cut the code out of the main program and had it print to the
floating point accuracy at the beginning of the loop and found this:
0.100000000000000005551115123126
0.200000000000000011102230246252
0.300000000000000044408920985006
0.400000000000000022204460492503
0.500000000000000000000000000000
0.599999999999999977795539507497
0.699999999999999955591079014994
0.799999999999999933386618522491
0.899999999999999911182158029987
0.949999999999999955591079014994
0.959999999999999964472863211995
0.969999999999999973354647408996

So I now see the reason why the < 0.8 comparison failed because the
value of $eta is actually 0.7999999 and not 0.8. I know there are fixes
for this such as making the comparions with 0.79, using a sprintf,
rounding, truncating, et cetera; but my officemate checked this code on
his machine and we got the same result.

My question is... WHY? Why is the addition below 0.5 of a number which
is slightly above 0.1, but then above 0.5 the addition becomes
something which is slightly less than 0.1? And, how does one fix (if
that's possible) this problem?

Thanks.

 

[Paul Lalli]

My question is... WHY? Why is the addition below 0.5 of a number which
is slightly above 0.1, but then above 0.5 the addition becomes
something which is slightly less than 0.1? And, how does one fix (if
that's possible) this problem?

This is not specific to Perl, but is a problem with computers in
general. Please read:
perldoc -q 999
perldoc perlnumber

Paul Lalli

 

[usenet]

So I now see the reason why the < 0.8 comparison failed because the
value of $eta is actually 0.7999999 and not 0.8.

perldoc -q 9999

 

[Charlton Wilbur]

k> My question is... WHY? Why is the addition below 0.5 of a
k> number which is slightly above 0.1, but then above 0.5 the
k> addition becomes something which is slightly less than 0.1?

perldoc -q 'long decimal'
http://docs.sun.com/source/806-3568/ncg_goldberg.html

k> And, how does one fix (if that's possible) this problem?

Start by understanding why and how it happens.

Charlton

 

[Ken]

Thank you all very much.

The fix is in and everything works fine.

 

[Tad McClellan]

I was digging through some code this morning after encountering a
program error and I came across the following...

For this bit of code:
#-------------------------------------------------------#
$eta = 0.5;
$etacrit = 0.97;
while ($eta <= $etacrit) {
($eta < 0.80) ? ($etastep = 0.1) :
($eta < 0.95) ? ($etastep = 0.05) :
($etastep = 0.01);
$eta += $etastep;
}
#-------------------------------------------------------#

Which *should* output:

No it shouldn't.

The code contains no statements that should make output,
so it should output nothing.