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From: Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups: comp.lang.javascript
Subject: Re: IE and IEE 754
Date: Sun, 31 Aug 2014 01:04:19 +0100
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Dr J R Stockton <reply1400@merlyn.demon.co.uk.invalid> writes:

> In comp.lang.javascript message <871ts0md0v.fsf@bsb.me.uk>, Fri, 29 Aug
> 2014 00:04:48, Ben Bacarisse <ben.usenet@bsb.me.uk> posted:
>
>>Gene Wirchenko <genew@telus.net> writes:
>>
>>> On Thu, 28 Aug 2014 11:27:30 +0100, Ben Bacarisse
>>> <ben.usenet@bsb.me.uk> wrote:
>>>
>>> [snip]
>>>
>>>>I think it must.  The ECMAScript specification imports IEEE 754's strict
>>>>rounding and representation rules.  The IEEE FP value closest to 123.45,
>>>>when multiplies by 100 must yield 12345 according to those rules.
>>>
>>>      Why?  What if the value that you have (and want) happens to be
>>> the value closest to 123.45?  When you multiply that value by 100,
>>> wouldn't you want 100 times the value instead of 12345?
>>
>>Yes, but you won't get it.  You'll get the nearest floating point
>>number, which is 12345.  Am I missing your point?
>
> With FF31, or IE8, or Opera 12.17, or Chrome 36.0, in WinXP sp3, FWIW,
> try
>
>         S = ""
>         for (J=1 ; J<111 ; J+=2) S += "\n" + J + "\t" +(1/J)*J
>         alert(S)
>
> And, in FF31 and Chrome 37.0,
>         (123.45/100)*100 - 123.45    ->    -1.4210854715202004e-14
>
> Two nominally-inverse arithmetic operations on IEEE Doubles do not
> always yield the exact input.

Of course not (I hope I never suggested otherwise).  I am happy to see
that the result you report is the one expected from IEEE 745's rules, so
I am not surprised that it's what you get in a wide range of browsers.
(In fact those that did not give this result are the ones I'd won't to
know about.)

-- 
Ben.