[for (t=[0:step:1-step]) ...] yields one less than expected

[charlieb]

I'm guessing here but it seems like the intent of the 1-step end condition in the following is that we exclude 1.0 from the array. Thus:

step = 0.01
echo([for (t=[0:step:1-step]) t])

would print [0 ... 0.99] i.e. 100 values but it doesn't. It prints only 99 values: [0 ... 0.98].
This is a little confusing when trying to adapt the code because I end up with a bunch of -2s where I would expect to find only -1s.
This is not exactly a big deal but if someone would confirm the intent is as I expect I will create a pull request with the fixes.

[charlieb]

Well this is a little odd. I guess this is an openscad bug:

 step = 0.01;
 echo(1.0/step);        

ECHO: 100

 echo([for (t=[0:1.0:1/step]) t]);    

ECHO: [0, ... 100]

 echo([for (t=[0:1.0:0.999999999/step]) t]);                            

ECHO: [0, ... 99]

 echo([for (t=[0:step:1]) t]);

ECHO: [0, ... 0.99]

I'm not sure what to do with that now. I guess it's a floating point issue but my original point still stands, kinda. Please advise.

[MichaelAtOz]

wiki

You should avoid step values that cannot be represented exactly as binary floating point numbers. Integers are okay, as are fractional values whose denominator is a power of two. For example, 0.25 (1/4) and 0.125 (1/8) are safe, but 0.2 (2/10) should be avoided. The problem with these step values is that your range may have too many or too few elements, due to inexact arithmetic.

[MichaelAtOz]

I didn't write it but the intent is probably as you say to exclude the last.
Where is that line from, I'm presuming it is in list-comprehension-demo?

[MichaelAtOz]

for (t=[0:step:1-step/2]) ensures that the final element is >= (1-step) and < 1

[MichaelAtOz]

This is a good primer to this specific floating point problem.
0.01 as a double is 0.01000000000000000020816681711721685132943093776702880859375

[nophead]

I avoid this problem by passing the number of steps rather than the step size. I then iterate with integers and divide by the the number of steps. For example: - function bezier_path(v, steps = 100) = // Returns a Bezier path from control points v with steps segments [for(i = [0 : steps], t = i / steps) bezier(t, v)]; Note that this does 101 iterations because a curve with 100 segments has 101 points.

…

On Tue, 30 Jul 2019 at 06:19, Michael ***@***.***> wrote: This is a good primer <https://www.exploringbinary.com/why-0-point-1-does-not-exist-in-floating-point/> to this specific floating point problem. 0.01 as a double is 0.01000000000000000020816681711721685132943093776702880859375 — You are receiving this because you are subscribed to this thread. Reply to this email directly, view it on GitHub <#13?email_source=notifications&email_token=AAEKHBKGRKX4W5GK2JNMW4DQB7FPDA5CNFSM4IHYE7G2YY3PNVWWK3TUL52HS4DFVREXG43VMVBW63LNMVXHJKTDN5WW2ZLOORPWSZGOD3CZRPA#issuecomment-516266172>, or mute the thread <https://github.com/notifications/unsubscribe-auth/AAEKHBISRMTG55RFYXVWI73QB7FPDANCNFSM4IHYE7GQ> .