Bug in std::floor?

[Vincent Lefevre]

On 2017-11-09 10:01:45 +0100, Mason wrote:
> > On 08/11/17 23:13 +0200, Nil Geisweiller wrote:
> >
> >> The following
> >>
> >> ---------------------------------------------------------------------
> >> #include <iostream>
> >> #include <cmath>
> >>
> >> int main()
> >> {
> >>    double v = 4.6;
> >>    std::cout << "v = " << v << std::endl;
> >>    std::cout << "v*100 = " << v*100 << std::endl;
> >>    std::cout << "floor(v*100) = " << std::floor(v*100) << std::endl;
> >> }
> >> ---------------------------------------------------------------------
> >>
> >> outputs
> >>
> >> ------------------
> >> v = 4.6
> >> v*100 = 460
> >> floor(v*100) = 459
> >> ------------------
> >>
> >> It that a bug?
> 
> There is a boring and excruciatingly long paper floating (ha!) around:
> "What Every Computer Scientist Should Know About Floating-Point Arithmetic"
> https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html
> 
> This other site tries to be more accessible:
> http://floating-point-gui.de/
> 
> The short version is that it is not possible to represent 0.1
> (or any multiple thereof) in (fractional) base 2.

Actually, that's mainly a language design bug, which doesn't show
the error for 460 (which is representable exactly).

> Basically, using "IEEE 754 double-precision binary floating-point format"
> 4.6 is approximated as D=4.5999999999999996447286321199499070644378662109375
> ( https://en.wikipedia.org/wiki/Double-precision_floating-point_format )
> 
> Thus, it is obvious why floor(D*100) equals 459.

No, this is not obvious. The multiplication by 100 introduces a
rounding error. Thus this doesn't prove that the rounded value
D*100 is strictly less than 460.

-- 
Vincent Lefèvre <vincent@vinc17.net> - Web: <https://www.vinc17.net/>
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