CODEWITHSUNDEEP
delphi-xe2
Jeff Cope
Jeff Cope

Reputation: 791

Why is Delphi XE2 calculating double values incorrectly?

I'm getting an invalid calculation result when using the double type in Delphi XE2. Below is code to a very simple calculation that should result in zero. Instead, it is returning -1.13686837721616E-13 which is incorrect. 

var
  dblValue1, dblValue2, dblValue3, dblTotal: double;
begin
  try
    dblValue1 := 1671.37;
    dblValue2 := 871.37;
    dblValue3 := 800.00;

    dblTotal := (dblValue1 - dblValue3);
    dblTotal := (dblTotal - dblValue2);
    Write(FloatToStr(dblTotal));
    ReadLn;
  except
    on E: Exception do
      Writeln(E.ClassName, ': ', E.Message);
  end;

This fails the same way with either a 32 or 64 bit build.

Why does this calculation fail?

Upvotes: 0

Views: 764

Answers (2)

William McBrine
William McBrine

Reputation: 2256

It doesn't fail -- you've just run into the limits of precision that are inherent in any representation of floating point arithmetic on a computer. For comparison, I just tried the same subtraction in an interactive Python shell:

>>> a = 1671.37
>>> b = 871.37
>>> a - b
799.9999999999999

The solution is to either accept the limits of floating point math, and round off your answers to a desired precision; or, switch to a different kind of numerical representation. For instance, since these values use two decimal places (dollars and cents?), you could multiply them by 100, and use integers (dividing by 100 only for display purposes, if and when necessary). Then there's binary-coded decimal, and arbitrary precision... I don't know offhand how you'd do those in Delphi, but they're probably overkill, depending on your application.

Upvotes: 2

therealmarv
therealmarv

Reputation: 3742

Well I'm getting exactly the same result in Python ;)

Python 2.7.6 (default, Feb 24 2014, 16:00:34)
[GCC 4.2.1 Compatible Apple LLVM 5.0 (clang-500.2.79)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> a = 1671.37
>>> b = 871.37
>>> c = 800.00
>>> t = a - c
>>> t = t - b
>>> print t
-1.13686837722e-13
>>>

Floating point arithmetic is not 100% precise (I just know it is complex to explain how a floating point number is represented inside the memory).

Please read this answer here:

https://stackoverflow.com/a/2080642/756056

Upvotes: 1

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