Re^9: test fails on 64bit uselongdouble Perl

Rounding to the nearest integers gives an error of +/- 1.0 even if the actual error is infinitesimally small.

I don't believe that. Care to explain?

You do a calculation, and the result is an integer $n. Due to precision limits you get instead $n+$eps, where abs($eps) is small.

Why wouldn't proper rounding return $n, but $n+1 or $n-1? At least that's how I understood your statement that it gives an error of +/- 1.0, which doesn't make sense to me at all.

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Re^10: test fails on 64bit uselongdouble Perl by ikegami (Patriarch) on Oct 30, 2009 at 15:01 UTC
You do a calculation, and the result is an integer $n. Number $n. You haven't rounded to an integer yet. `$n = 5.5; $eps = -0.00000000000001; printf "%.0f\n", $n; # 6 expected printf "%.0f\n", $n+$eps; # 5 get` [download] If his calculations are suppose to always give integers before rounding, then yes, there's no problem.	[reply] [d/l]