MOD operator problems

Hi, All,
I'm intermediate for VB6. I want to get remainder from divided two doubles. I used ,

Dim a As Double, b As Double, result As Double
b = 8333.33
a = 58333.31

result = a - (b * Fix(a / b))

result is 8333.33; Why? Please help.

Code:

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a=58333,31    b=8333,33
Fix(a / b) ->  6
b * 6 ->  49999,98
a - (49999,98) ->  8333,33

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What's your actual question?

https://stackoverflow.com/questions/...dulo-operation

Code:

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The algorithm you want, to limit a floating point value between 0 and some modulus n:

Double fmod(Double value, Double modulus)
{
    return value - Trunc(value/modulus)*modulus;
}
for example pi mod e (3.14159265358979 mod 2.718281828459045)

3.14159265358979 / 2.718281828459045
  = 1.1557273497909217179

Trunc(1.1557273497909217179)
  = 1

1.1557273497909217179 - 1
  = 0.1557273497909217179

0.1557273497909217179 * e
  = 0.1557273497909217179 * 2.718281828459045
  = 0.42331082513074800
pi mod e = 0.42331082513074800

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VB6:

Code:

---

Private Sub Form_Load()
  Dim a As Double, b As Double, result As Double

  b = 8333.33
  a = 58333.31

  result = a - (b * Fix(a / b))
  Debug.Print result
  Debug.Print fmod(a, b)

  Debug.Print fmod(3.14159265358979, 2.71828182845905)
  ' -> 0.423310825130745
End Sub

Private Function fmod(value As Double, modulus As Double) As Double
    fmod = value - Fix(value / modulus) * modulus
End Function

---

b = 8333.33
a = 58333.31

a/b =7 and no remain. So Fix(a / b) should be 7. But Fix(a / b)=6, is it correct?

That's the problem with floating point number respresentation, calculation and the internal storage in 64 bits.

Code:

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? 7 * 8333.33 - 58333.31
 1.81898940354586E-12

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Quote:

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Originally Posted by Arnoutdv

That's the problem with floating point number respresentation, calculation and the internal storage in 64 bits.

Code:

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? 7 * 8333.33 - 58333.31
 1.81898940354586E-12

---

---

Yep, a floatingpoint-mod-function needs to take these "small epsilons" into account:

Code:

---

Private Sub Form_Load()
  Dim a#: a = 58333.31
  Dim b#: b = 8333.33
 
  Debug.Print a - b * Fix(a / b), fmod(a, b)
End Sub

Public Function fmod(ByVal Dividend#, ByVal Divisor#) As Double
  fmod = Dividend - Fix(Dividend / Divisor) * Divisor
 
  If Abs(fmod - Divisor) < Abs(Divisor * 0.000000001) Then fmod = 0
End Function

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HTH

Olaf

Or use some small EPSILON value to fix the Fix function and to implemented comparison to zero with |A| < EPSION using Abs function like this:

Code:

---

Option Explicit

Private Const EPSILON As Double = 0.0000001

Private Sub Form_Load()
    Dim a#: a = 58333.31
    Dim b#: b = -8333.33
 
    Debug.Print "No remainder? -> "; Abs(fmod(a, b)) < EPSILON
    Debug.Print Format$(fmod(a, b), "0.0000")
    Debug.Print Format$(fmod(a, b), "0.00000000")
    Debug.Print Format$(fmod(a, b), "0.000000000000")
End Sub

Private Function fmod(value As Double, modulus As Double) As Double
    fmod = value - Fix(value / modulus + Sgn(value) * Sgn(modulus) * EPSILON) * modulus
End Function

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cheers,
</wqw>

Quote:

---

Originally Posted by wqweto

Or use some small EPSILON value to fix the Fix function and to implemented comparison to zero with |A| < EPSION using Abs function like this:

Code:

---

Option Explicit

Private Const EPSILON As Double = 0.0000001

Private Sub Form_Load()
    Dim a#: a = 58333.31
    Dim b#: b = -8333.33
 
    Debug.Print "No remainder? -> "; Abs(fmod(a, b)) < EPSILON
    Debug.Print Format$(fmod(a, b), "0.0000")
    Debug.Print Format$(fmod(a, b), "0.00000000")
    Debug.Print Format$(fmod(a, b), "0.000000000000")
End Sub

Private Function fmod(value As Double, modulus As Double) As Double
    fmod = value - Fix(value / modulus + Sgn(value) * Sgn(modulus) * EPSILON) * modulus
End Function

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cheers,
</wqw>

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Thanks for the indirect hint ...
(regarding the fmod-function I've posted in #6, which fails to return a zero with that negative Divisor-input).

Here's a correction:

Code:

---

Public Function fmod(ByVal Dividend#, ByVal Divisor#) As Double
  Const EPS# = 0.00000001
  If Divisor < 0 Then Divisor = -Divisor
  If Dividend > 0 Then
    fmod = Dividend - Int(Dividend / Divisor) * Divisor
    If Abs(fmod - Divisor) < Divisor * EPS Then fmod = 0
  Else
    fmod = Dividend + Int(-Dividend / Divisor) * Divisor
    If Abs(fmod + Divisor) < Divisor * EPS Then fmod = 0
  End If
End Function

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Quite a bit longer than the first version (or yours) - but it now has less internal function-calls - and is therefore about 25% faster in native-compiled binaries... (though still about 10% slower in IDE or PCode).

Olaf