my $r = ($i)*(1e-6);    #radii are in microns

Why do you put both factors in brackets?

Integers will be presented without any loss of precision in Perl

But only as long as the number "fits" into perls integer type, which typically is a 32 bit signed integer or 64 bit signed integer. Once you exceed the limit (+/- 231 resp. +/- 263), perl silently converts the value to a floating point number with loss of precision (perlnumber).

Alexander

--
Today I will gladly share my knowledge and experience, for there are no sweeter words than "I told you so". ;-)

In reply to Re^2: Improve my Code: Subroutines by afoken
in thread Improve my Code: Subroutines by cheech

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