I'm sure there are more elegant ways to do what you want, but here is one thought:

» 3.3-2.2-1.1

ans =

 -4.4409e-016

» multiplier = 1e9

multiplier =

  1.0000e+009

» (multiplier*3.3-multiplier*2.2-multiplier*1.1)/multiplier

ans =

     0
 

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Thank you very much electricpete. Your solution works, but I prefer a more elegant one, because it is going to be a application to distribute. Thus, the operation that I do where I have that problem is more complex than 3.3-2.2-1.1, although it is the same problem, so if I use your solution the code would be very very more complex, something I am avoiding.

I would like some solution as easy as:

double (3.3)-double (2.2)-double (1.1)=0

But this does not work of course.

Thank you very much.

In the other hand my question is: Should I trust in an application that gets this output??:

floor (3.3-2.2-1.1)

ans =

    -1

I think it is a basic operation, and it makes it wrong.

There are discussions like this at least once a week in the Matlab USENET newsgroup (comp.soft-sys.matlab).  There is a FAQ devoted to it:

http://matlabwiki.mathworks.com/MATLAB_FAQ#Why_is_0.3-0.2-0.1_not_equal_to_zero_.28or_similar.29.3F

Simply put, binary fractions (as used by IEEE 754 floating point representation) can't hold multiples of 1/10 (1/100, 1/1000, etc) in an exact way, so there is always a rounding error when working with these numbers.  You just can't write code that relies on exact representation unless it's symbolic.

As an aside, most calculators use BCD (Binary coded decimal) which wastes memory by storing each decimal digit in a 4 byte nibble.

- Steve

Thank you SomptingGuy,

It seems there is no a basic solution to this problem, so I will try to write my code in a different way, where it could be possible to store the values in different format to double, or using the round () function.

Thank you very much.

Sorry to sound negative, but round() won't help you either.  You just can't trim those pesky unwanted digits - they come back.   You need to change the way you compare values, assuming that's the problem.

if (a==b)

should be replaced by something like

if(abs(a-b)<tol)

where tol is some small number, possibly based on eps if you want to be rigorous.

- Steve

Alternatively implement BCD arithmetic in Matlab. I haven't done it, but can't really see it as being difficult, just slow.

 

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Ack!! a four bit nibble.  I do have an 'O' level in computer studies.

- Steve

How about this (needs symbolic toolbox):
» vpa(3.3)-vpa(2.2)-vpa(1.1)
 
ans =
 
0

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Also one has to think about the precision of the numbers that you work with.  A 1E-16 error in treating numbers on the order of magnitude of 1 wouldn't be important unless your numbers have 16 significant figures to begin with.
 

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I can't think of a case where it has come up, but if you were calculating the exponent of something then

x^2

and x^2.0000000000000000000000000000001

have very different properties. That is a quick example, perhaps there are others.

Cheers

Greg Locock

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Like I said... "unless your numbers have 16 significant figures to begin with".  You are assuming this 2 has 16+ significant figures and if so the unless clause is applicable.  

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I was wrong on two counts, #sf, and in fact the example is unhelpful, the difference between the two is tiny.

Cheers

Greg Locock

SIG:Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of Eng-Tips.

Fractions cannot be represented exactly in a binary floating point system, unless they are a power of 1/2.  While there are problems that thrive on such roundoff errors, most real world problems can get along quite fine.

In your case, if this is a really big deal, then you need to code the problems directly in fractional representations, or use more significant figures, but that just pushes the problem further out in the sig-fig realm.  There are various mathmatics applications, like Mathcad, or Mathematica, that can take various exact representations, and produce exact answers.

TTFN

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