If you really have a loop and you want to avoid the accumulation of error, you can avoid use an integer as the loop counter, and compute the current value of the real from the integer. 

# Loop over [0.00..1.00] by 0.05
for (my $i = 0; $i <= 100; $i += 5) {
   my $x = $i / 100;
   printf("[%d] x= %g (%20.40f)\n", ++$c, $x, $x);
}

[download]
# Loop over [0.00..1.00] by 0.05
for (my $i = 0; $i <= 20; $i++) {
   my $x = $i / 20;
   printf("[%d] x= %g (%20.40f)\n", $i+1, $x, $x);
}

[download]
# Loop over [0.00..1.00] by 0.05
for (0..20) {
   my $x = $_ / 20;
   printf("[%d] x= %g (%20.40f)\n", $_+1, $x, $x);
}

[download]

All of the above output the following: 

[1] x= 0 (0.0000000000000000000000000000000000000000)
[2] x= 0.05 (0.0500000000000000030000000000000000000000)
[3] x= 0.1 (0.1000000000000000100000000000000000000000)
[4] x= 0.15 (0.1499999999999999900000000000000000000000)
[5] x= 0.2 (0.2000000000000000100000000000000000000000)
[6] x= 0.25 (0.2500000000000000000000000000000000000000)
[7] x= 0.3 (0.2999999999999999900000000000000000000000)
[8] x= 0.35 (0.3499999999999999800000000000000000000000)
[9] x= 0.4 (0.4000000000000000200000000000000000000000)
[10] x= 0.45 (0.4500000000000000100000000000000000000000)
[11] x= 0.5 (0.5000000000000000000000000000000000000000)
[12] x= 0.55 (0.5500000000000000400000000000000000000000)
[13] x= 0.6 (0.5999999999999999800000000000000000000000)
[14] x= 0.65 (0.6500000000000000200000000000000000000000)
[15] x= 0.7 (0.6999999999999999600000000000000000000000)
[16] x= 0.75 (0.7500000000000000000000000000000000000000)
[17] x= 0.8 (0.8000000000000000400000000000000000000000)
[18] x= 0.85 (0.8499999999999999800000000000000000000000)
[19] x= 0.9 (0.9000000000000000200000000000000000000000)
[20] x= 0.95 (0.9499999999999999600000000000000000000000)
[21] x= 1 (1.0000000000000000000000000000000000000000)

[download]

In reply to Re: adding numbers and floating point errors by ikegami
in thread adding numbers and floating point errors by smeenz

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