FAQ 4.4 Does Perl have a round() function? What about ceil() and floor()? Trig functions...

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4.4: Does Perl have a round() function?  What about ceil() and floor()?  Trig

    Remember that "int()" merely truncates toward 0. For rounding to a
    certain number of digits, "sprintf()" or "printf()" is usually the
    easiest route.

            printf("%.3f", 3.1415926535);   # prints 3.142

    The "POSIX" module (part of the standard Perl distribution) implements
    "ceil()", "floor()", and a number of other mathematical and
    trigonometric functions.

            use POSIX;
            $ceil   = ceil(3.5);   # 4
            $floor  = floor(3.5);  # 3

    In 5.000 to 5.003 perls, trigonometry was done in the "Math::Complex"
    module. With 5.004, the "Math::Trig" module (part of the standard Perl
    distribution) implements the trigonometric functions. Internally it uses
    the "Math::Complex" module and some functions can break out from the
    real axis into the complex plane, for example the inverse sine of 2.

    Rounding in financial applications can have serious implications, and
    the rounding method used should be specified precisely. In these cases,
    it probably pays not to trust whichever system rounding is being used by
    Perl, but to instead implement the rounding function you need yourself.

    To see why, notice how you'll still have an issue on half-way-point

            for ($i = 0; $i < 1.01; $i += 0.05) { printf "%.1f ",$i}

            0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7
            0.8 0.8 0.9 0.9 1.0 1.0

    Don't blame Perl. It's the same as in C. IEEE says we have to do this.
    Perl numbers whose absolute values are integers under 2**31 (on 32 bit
    machines) will work pretty much like mathematical integers. Other
    numbers are not guaranteed.


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Re: FAQ 4.4 Does Perl have a round() function? What about ceil() and floor()? Trig functions?

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That example conflates two different issues:

1) 0.05 is not exactly representable as a binary floating point number,
   so $i doesn't get the values 0.00, 0.05, 0.10, 0.15, 0.20, 0.25, ...
   as one might naively expect. Instead it gets the values
   (rounded to 20 digits after the comma). So the second value
   (0.05000000000000000278) is actually a bit above 0.05 and therefore
   has to be rounded up according to the "round to nearest" rule.
   So most of these numbers are not at the "half-way-point", and can be
   clearly and anambiguosly rounded (but the result is not what you
   expect when you think they are at the half-way-point.

2) The one exception is 0.25 which just happens to be exact because all
   the rounding errors cancel out at that point[1]. 0.25 is exactly
   half-way between 0.2 and 0.3 so "round to nearest" cannot decide
   between these two. In school we all learned to round up in this case.
   The IEEE rules, however, mandate that one must round to the nearest
   *even* number, i.e., 0.2 in this case.

I'm not sure which of these issues the example should demonstrate, but
showing both in a single example without adequate explanation is
confusing. The "half-way-point" problem can be demonstrated correctly

            for ($i = 0; $i <= 10; $i += 0.5) { printf "%.0f ",$i}

            0 0 1 2 2 2 3 4 4 4 5 6 6 6 7 8 8 8 9 10 10

The fact that decimal fractions are not generally representable in
binary fp and that adding them is an especially bad idea can be
demonstrated better with:

        for ($i = 0, $j = 0; $i < 1.01; $i += 0.05, $j += 5) {
        print $i * 100 - $j, " ";

        0 0 0 1.77635683940025e-15 0 0 0 0 0 -7.105427357601e-15
        -7.105427357601e-15 -7.105427357601e-15 0 0 0
        1.4210854715202e-14 1.4210854715202e-14 1.4210854715202e-14
        2.8421709430404e-14 2.8421709430404e-14 2.8421709430404e-14

(Although there are still too many 0s in the output - maybe somebody can
find a better example).

Quoted text here. Click to load it

2**53 actually, if the machine uses IEEE FP arithmetic.


[1] I didn't expect that and I even repeated the calculation with pencil
and paper. Adding a number which is slightly larger than 0.05 5 times
gives exactly 0.25 in IEEE double arithmetic.

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