Where did my last loop go? - perl

Question

In the course of updating some legacy Perl code I came across a situation where a C-style for loop terminates prematurely.
The Problem
The situation can be recreated as follows (same result for me on Linux and Windows):
$ perl
for($i = 0.8; $i <= 2.5; $i+=0.1){
print $i, "\n";
}
__END__
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
The 2.5 is noticeably absent; it should've passed the $i <= 2.5 conditional.
The Fix
Realizing this may be due to floating-point representations, as outlined in perldoc perlfaq4, I updated the conditional:
$ perl
for($i = 0.8; sprintf( '%.1f', $i ) <= 2.5; $i += 0.1 ) {
print $i, "\n";
}
__END__
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5 <--- Gotcha !
And service was resumed (or so I thought).
The Verification
I wanted to prove that the final $i was indeed greater than 2.5, so I wrapped the increment in a do loop:
$ perl
for($i = 0.8; $i <= 2.5; do { $i += 0.1; print $i, "\n"; } ) {}
__END__
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5 <--- Surprise!
The test results seem to indicate that floating-point representation is not the issue here.
What is going on here?

Answer 1

You're suprised that a value greater than 2.5 printed as 2.5? Stop being surprised, that's normal floating point stuff. A plain print doesn't show all the digits. Try this:
printf "%.66g\n", $i
in place of your print statement. On my machine the final value printed is 2.5000000000000013. Yours will probably be similar.

Answer 2

Maybe it is because of float, it is not just as 2.5, you can use it as follows:
for ($i = 0.8; $i < 2.6; $i += 0.1) {
or you can see:
for ($i = 0.8; $i <= 2.5; $i += 0.1) {
print $i, "\n";
}
print $i, "\n";
print $i <= 2.5 ? 'TRUE' : 'FALSE';
output:
0.8
0.9
1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
FALSE

Answer 3

Computer use a binary storage, not a decimal! If you store the decimal number 0.1 as a binary floating point number, it will result in an infinte number ob binary digits:
dezimal 0.1 = binary 0.000110011001100110011001100110011...
So if you don't use integers, you will get rounding-errors. The dezimal value 0.1 might be stored as the digital number 0.0001100110011001101 and this is dezimal 0.100006104
Adding 25 times 0.100006104 gives 2.50015259 and this is more than 2.5, so this last value is out of range.

Answer 4

The value of i is being increased at different times. Notice your first value in the last run is 0.9, not 0.8.
The problem with the first one is floating point, try using 2.5001 in the condition. Try it using integers, 8 to 25 by 1, the <= works fine.
for($i = 8; $i <= 25; $i += 1 ) {
print $i/10, "\n";
}
Or
print $_/10, "\n" for 8..25;
or
my $epsilon = .00000001;
for(my $i = 0.8; ($i - 2.5) <= $epsilon; $i += .1 ) {
print $i, $/;
}