For my machine, it will give the wrong answer for any time $yearlyamount equals k * 13, for k at least 9. This is because 52 is 4 * 13, and hence the division by 52 gives the exact answer, and subtracting 0.005 (which cannot be represented exactly in binary) from it causes %f to round it down to the next decimal. For instance, 117 / 52 - 0.005, on my computer with my Perl (5.8.6, with longdoubles and 64bitints) is about 2.244999999999999999895916591441391574335284531116485595703125. Just a tiny, tiny bit below 2.245, but that's enough to give the answer 2.24, instead of 2.25.

Doing arithmetic with reals is very tricky, and it's therefore better to use a good module. POSIX.pm for instance, with its floor method.

In reply to Re^2: Rounding of a number... by Anonymous Monk
in thread Rounding of a number... by mkenney

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