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.

Perl 6 - links to (nearly) everything that is Perl 6.

In reply to Re^9: test fails on 64bit uselongdouble Perl by moritz
in thread test fails on 64bit uselongdouble Perl by frankcox

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