Just to follow up on this topic, here's what Matlab does:
2*10.97 + 23.9985
ans =
45.9385
2*10.97 + 23.9985 == 45.9385
ans =
0
format long
2*10.97 + 23.9985
ans =
45.938500000000005
2*10.97 + 23.9985 == 45.938500000000005
ans =
1
Note that Matlab's display value is 45.9385, but the actual result is
45.938500000000005. You only see the display value if you set the
display to show all digits ("format long")
Stuart
On Wed, 5 Nov 2014, K Leo wrote:
julia> 2*10.97 + 23.9985
45.938500000000005
julia> 2*10.97 + 23.9985 == 45.938500000000005
true
Amazing. I never expected this. Is floating point comparison going to be
guaranteed?
On 2014?11?05? 08:48, Stefan Karpinski wrote:
Some systems round their answers as John said but it's easy to check that
it's a lie:
R version 3.1.0 (2014-04-10) -- "Spring Dance"
> 2*10.97 + 23.9985
[1] 45.9385
> 2*10.97 + 23.9985 == 45.9385
[1] FALSE
This is perl 5, version 16, subversion 2 (v5.16.2)
DB<1> x 2*10.97 + 23.9985
0 45.9385
DB<2> x 2*10.97 + 23.9985 == 45.9385
0 ''
I don't have a working copy of Matlab right now, but I think it does this
too.
On Tue, Nov 4, 2014 at 8:31 PM, Neil Devadasan <[email protected]
<mailto:[email protected]>> wrote:
Thanks
On Tuesday, November 4, 2014 2:13:37 PM UTC-5, John Myles White
wrote:
Hi Neil,
Julie does math the same way that all computers do math.
You're probably coming from another language where a lot of
effort is invested into pretending that computers offer a
closer approximation to abstract mathematics than they
actually do. Those systems have been lying to you.
Put another way: you just took the red pill by using Julia.
-- John
On Nov 4, 2014, at 11:06 AM, Neil Devadasan
<[email protected]> wrote:
> julia> f(x::Float64, y::Float64) = 2x + y;
>
> julia> f(10.97,23.9985)
> 45.938500000000005
>
> The above method execution of function f returns an answer
that I cannot understand. Can someone clarify?
>
> Thank you.
