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    SubjectRe: [PATCH] Add "is_power_of_2" checking to log2.h.
    On Wed, 31 Jan 2007, Vegard Nossum wrote:

    > On Tue, January 30, 2007 3:12 pm, Jan Engelhardt wrote:
    > >
    > >>> Why the qualifier? Zero *is* not a power of 2, is it?
    > >
    > > No, it is not:
    > >
    > > In[1]:= Solve[2^n == 0, n]
    > >
    > > Out[1]= {}
    > >
    > > So says Mathematica5.
    > As a side note, I would just like to point out that Mathematica does
    > not deal with modular arithmetic by default (which programmers very
    > much do). In fact, in modular arithmetic, zero IS a power of two.
    > 2^n = 0 (mod 2^n)
    > To see if it holds for bytes, substitute n = 8, and you get 2^8 = 0
    > (mod 256). In other words: Zero is the eighth power of two modulo
    > 256.

    that's a bit esoteric but, yes, you make a good point.

    > Modular arithmetic is, however, very often a source of errors in
    > programming (unchecked-for overflows and underflows), and it is
    > questionable whether the programmer would really want 0 reported as
    > a power of two.

    precisely. given the definition of "is_power_of_2()" that's been
    published, some people will (quite correctly) point out that using
    that macro in place of the test "(n & (n - 1))" isn't quite the same
    thing. and they would be correct.

    but one wonders how many programmers have been using that very test
    "((n & (n - 1))" without realizing that it would accept zero, and that
    it has been accepting zero all this time, perhaps leading to weird and
    inexplicable errors.

    rewriting that test as "is_power_of_2()" may, in fact, cause some
    things to suddenly break, but perhaps those are things that should
    have been *forced* to break anyway, to identify where the condition
    check was incorrect all this time, and someone's just been lucky, or
    something like that.

    ***NOTE***: and that's why i suggested that the individual
    maintainers might want to make this substitution themselves, to make
    sure everything continues to work. because if you leave it as a
    project for the kernel janitors, they might not realize what you had
    in mind. so it's in your best interest to start cleaning this up on
    your own.

    just my $0.02 (Cdn.)


    Robert P. J. Day
    Linux Consulting, Training and Annoying Kernel Pedantry
    Waterloo, Ontario, CANADA
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