CODEWITHSUNDEEP
vbams-accessdoublemodulus
Blackhawk
Blackhawk

Reputation: 6120

Mod with Doubles

Am I doing something wrong or does the VBA Mod operator actually not work with floating point values like Doubles?

So I've always sort of assumed that the VBA Mod operator would work with Doubles based on the VB documentation, but in trying to figure out why my rounding function doesn't work, I found some unexpected Mod behavior.

Here is my code:

Public Function RoundUp(num As Double, Optional nearest As Double = 1)
    RoundUp = ((num \ nearest) - ((num Mod nearest) > 0)) * nearest
End Function

RoundUp(12.34) returns 12 instead of 13 so I dug a little deeper and found that:

12.5 Mod 1 returns 0 with the return type of Long, whereas I had expected 0.5 with a type of Double.

Conclusion

As @ckuhn203 points out in his answer, according to the VBA specification,

The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result.

And

Usually, the data type of result is a Byte, Byte variant, Integer, Integer variant, Long, or Variant containing a Long, regardless of whether or not result is a whole number. Any fractional portion is truncated.

For my purposes, I need a floating point modulo and so I have decided to use the following:

Public Function FMod(a As Double, b As Double) As Double
    FMod = a - Fix(a / b) * b

    'http://en.wikipedia.org/wiki/Machine_epsilon
    'Unfortunately, this function can only be accurate when `a / b` is outside [-2.22E-16,+2.22E-16]
    'Without this correction, FMod(.66, .06) = 5.55111512312578E-17 when it should be 0
    If FMod >= -2 ^ -52 And FMod <= 2 ^ -52 Then '+/- 2.22E-16
        FMod = 0
    End If
End Function

Here are some examples:

FMod(12.5, 1) = 0.5 FMod(5.3, 2) = 1.3 FMod(18.5, 4.2) = 1.7

Using this in my rounding function solves my particular issue.

Upvotes: 13

Views: 18735

Answers (5)

rom m
rom m

Reputation: 101

Public Function Modi(d as double) as double
Modi = d - Int(d)
End Function

Dim myDoule as Double
myDoule = 1.99999

Debug.Print Modi(myDoule)

0.99999

Upvotes: 0

user14632325
user14632325

Reputation: 1

Try This in VBS:

Option Explicit

Call Main()

Sub Main()

    WScript.Echo CStr(Is_Rest_Of_Divide_Equal_To_Zero(506.25, 1.5625))

End Sub


Function Is_Rest_Of_Divide_Equal_To_Zero(Divident, Divisor)

    Dim Result
    Dim DivideResult

    If Divident > Divisor Then
        DivideResult = Round(Divident/Divisor, 0)
        If (DivideResult * Divisor) > Divident Then
            Result = False
        ElseIf (DivideResult * Divisor) = Divident Then
            Result = True
        ElseIf (DivideResult * Divisor) < Divident Then
            Result = False
        End If
    ElseIf Divident = Divisor Then
        Result = True
    ElseIf Divident < Divisor Then
        Result = False
    End If

    Is_Rest_Of_Divide_Equal_To_Zero = Result

End Function

Upvotes: 0

Russell Munro
Russell Munro

Reputation: 563

As a work around your can do some simple math on the values. To get two decimal of precision just multiply the input values by 100 and then divide the result by 100.

result = (123.45*100 Mod 1*100)/100
result = (12345 Mod 100)/100
result = 0.45

I'm late to the party, but just incase this answer is still helpful to someone.

Upvotes: 2

RubberDuck
RubberDuck

Reputation: 12728

According to the VB6/VBA documentation

The modulus, or remainder, operator divides number1 by number2 (rounding floating-point numbers to integers) and returns only the remainder as result. For example, in the following expression, A (result) equals 5. A = 19 Mod 6.7 Usually, the data type of result is a Byte, Byte variant, Integer, Integer variant, Long, or Variant containing a Long, regardless of whether or not result is a whole number. Any fractional portion is truncated. However, if any expression is Null, result is Null. Any expression that is Empty is treated as 0.

Remember, mod returns the remainder of the division. Any integer mod 1 = 0. 

debug.print 12 mod 1 
'12/1 = 12 r 0

The real culprit here though is that vba truncates (rounds down) the double to an integer before performing the modulo.

?13 mod 10
 '==>3 
?12.5 mod 10
 '==>2 

debug.print 12.5 mod 1
'vba truncates 12.5 to 12
debug.print 12 mod 1
'==> 0

Upvotes: 10

Davis
Davis

Reputation: 250

I believe that the Mod operator calculates with long type only. The link that you provided is for VB.Net, which is not the same as the VBA you use in MSAccess.

The operator in VBA appears to accept a double type, but simply converts it to a long internally.

This test yielded a result of 1.

9 Mod 4.5

This test yielded a result of 0.

8 Mod 4.5

Upvotes: 4

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