Both 1/10 and 16/10 are periodic numbers in binary, just like 1/3 is in decimal. It would take an infinite amount of memory to store 0.1 and 1.6 as floats. Some rounding error occurs.
And to see what's happening for you:$ perl -e' printf "%.20f\n", 1.6;' 1.60000000000000008882
$ perl -e' for ($i=1.1;$i<=1.6;$i+=.1) { printf "%.20f\n", $i; } printf "\n%.20f\n", $i; ' 1.10000000000000008882 1.20000000000000017764 1.30000000000000026645 1.40000000000000035527 1.50000000000000044409 1.60000000000000053291
One solution is to use integers for the counter, and generate the floats from them
$ perl -le'for my $i (1..6) { print 1+$i/10 }' 1.1 1.2 1.3 1.4 1.5 1.6
It gives precise checks for the loop, and there's no accumulation of error for the result.
In reply to Re: Floating Point Looping In Perl
by ikegami
in thread Floating Point Looping In Perl
by kiruthika.bkite
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