Double data type numbers error

Question

1.00/5 (1 vote)

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When i try to cal in VB or C# or VBA
0.43 + 0.000001 - 0.430001
The result will not be zero! Why?
Please try it!

What I have tried:

VB

Sub Test()
    Dim V As Double
    V = 0.43 + 0.000001 - 0.430001
    If V <> 0 Then
        MsgBox "Why we goes hear?"
    End If
End Sub

Posted 27-Apr-16 5:58am

NewPast

Updated 27-Apr-16 6:35am

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Comments

ZurdoDev 27-Apr-16 12:05pm

The Precision of each number not the same so you get these weird issues.

RedDk 27-Apr-16 13:09pm

As all the solutions are correct, I'd pick the most right, which would be ppolymorphes. Because at the end of the day, all "relevanting" calculations are accumulatively absolute. Programmers do indeed use floating point even though they know they are accumulating error but using it consistantly throughout their system makes it ... ok.

phil.o 27-Apr-16 16:16pm

I updated my answer with interesting follow-ups and solution.

4 solutions

Solution 3

Use Decimal instead of Double.

If you have to use Double, avoid equality / inequality comparisons. Pick a precision for your application. As an example of six decimals precision, replace "if (x <> 0.0)" with "if ( Abs(x) < 0.000001)".

Posted 27-Apr-16 6:27am

bling

Updated 27-Apr-16 6:30am

v2

Solution 4

That is what floating point is about.
The downside of floating point is that values are not exact in most cases.
Thus all calcs are accumulating errors.

Posted 27-Apr-16 6:35am

Patrice T

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F-ES Sitecore · Accepted Answer · 2016-04-27T06:09:00

Solution 1

Doubles are an approximation of the numbers, not the exact numbers so they shouldn't be used where precision is required as you'll get these kinds of issues.

Posted 27-Apr-16 6:09am

F-ES Sitecore

phil.o · Accepted Answer · 2016-04-27T06:09:00

You are facing an issue with floating-point arithmetic precision.
Please have a look at What Every Computer Scientist Should Know About Floating-Point Arithmetic[^]

[Edit] By the way, there exists an excellent article from John Simmons / outlaw programmer here on CP; it presents a way to circumvent this problem in equality comparisons between floating-point values. There:
Reliable Floating Point Equality Comparison[^]

As the code in this article is for C++, here is the extension method I'm using in C# for all equality comparisons with floating-point valuetypes:

C#

public static bool AlmostEquals(this double value, double other, double epsilon = 1e-15)
{
   return (((other - epsilon) < value) && (value < (other + epsilon)));
}

It is based on the comment from B. Clay Shannon in the same article.

Usage of this snippet applied to your case could be:

VB

Sub Test()
    Dim V As Double
    V = 0.43 + 0.000001 - 0.430001
    If Not V.AlmostEquals(0) Then
        MsgBox "Why we goes hear?"
    End If
End Sub