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Re: 0.3*3 == 0.9?
From: |
Jaroslav Hajek |
Subject: |
Re: 0.3*3 == 0.9? |
Date: |
Wed, 25 Aug 2010 20:59:23 +0200 |
2010/8/25 Michael Creel <address@hidden>:
> 2010/8/25 Jordi Gutiérrez Hermoso <address@hidden>
>>
>> 2010/8/25 Jordi Gutiérrez Hermoso <address@hidden>:
>> > On 25 August 2010 09:13, Michael Creel <address@hidden> wrote:
>> >> octave:1> 0.3*2 == 0.6
>> >> ans = 1
>> >> octave:2> 0.3*3 == 0.9
>> >> ans = 0
>> >> octave:3>
>> >
>> > http://docs.sun.com/source/806-3568/ncg_goldberg.html
>>
>> Oops, sorry, I hit the "send" button too soon.
>>
>> Anyways, yes, this is completely normal. 0.3 has an infinite binary
>> expansion because its denominator has factors other than 2, so you'll
>> get a rounding error. Use something like abs(0.3*3 - 0.9) < 10*eps
>> instead or whatever seems like an appropriate approximation.
>
> Right, I'm aware of that. My question is whether or not Octave's comparison
> of floats takes it into account. My impression was that it does, so I was
> surprised by this.
Interesting. Where did you get that impression from and what did you
actually expect Octave to do?
--
RNDr. Jaroslav Hajek, PhD
computing expert & GNU Octave developer
Aeronautical Research and Test Institute (VZLU)
Prague, Czech Republic
url: www.highegg.matfyz.cz
- 0.3*3 == 0.9?, Michael Creel, 2010/08/25
- Re: 0.3*3 == 0.9?, Jordi Gutiérrez Hermoso, 2010/08/25
- Re: 0.3*3 == 0.9?, Jordi Gutiérrez Hermoso, 2010/08/25
- Re: 0.3*3 == 0.9?, Michael Creel, 2010/08/25
- Re: 0.3*3 == 0.9?, Jordi Gutiérrez Hermoso, 2010/08/25
- Re: 0.3*3 == 0.9?,
Jaroslav Hajek <=
- Re: 0.3*3 == 0.9?, Michael Creel, 2010/08/25
- Re: 0.3*3 == 0.9?, José Luis García Pallero, 2010/08/25
- Re: 0.3*3 == 0.9?, Jordi Gutiérrez Hermoso, 2010/08/25
- Re: 0.3*3 == 0.9?, Jaroslav Hajek, 2010/08/26
Re: 0.3*3 == 0.9?, Marc Weber, 2010/08/25
Re: 0.3*3 == 0.9?, CdeMills, 2010/08/26