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Re: Bug in std::floor?
- From: Mason <slash dot tmp at free dot fr>
- To: Nil Geisweiller <ngeiswei at googlemail dot com>
- Cc: Jonathan Wakely <jwakely at redhat dot com>, GCC help <gcc-help at gcc dot gnu dot org>
- Date: Thu, 9 Nov 2017 10:01:45 +0100
- Subject: Re: Bug in std::floor?
- Authentication-results: sourceware.org; auth=none
- References: <49c6843e-4596-fd46-7f38-ab75d4cc9794@gmail.com> <20171108222503.GA2558@redhat.com>
> 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.
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.
Regards.