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Author: jimmelanson

example_Modulus_Zero.xlsm

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+Attribute VB_Name = "udf_modulus_zero"
+Option Explicit
+
+Private Sub ModulusZero_Test()
+    Debug.Print "ModulusZero(3.001, 3) -> " & ModulusZero(3.001, 3)
+    Debug.Print "ModulusZero(2, 0.3) -> " & ModulusZero(2, 0.3)
+    Debug.Print "ModulusZero(3, 0.3) -> " & ModulusZero(3, 0.3)
+    Debug.Print "ModulusZero(6.3, 0.3) -> " & ModulusZero(6.3, 0.3)
+    Debug.Print "ModulusZero(57, 0.4) -> " & ModulusZero(57, 0.4)
+    Debug.Print "ModulusZero(6324692, 415) -> " & ModulusZero(6324692, 415)
+    Debug.Print "ModulusZero(6324600, 415) -> " & ModulusZero(6324600, 415)
+    Debug.Print "ModulusZero(6324600.99, 415) -> " & ModulusZero(6324600.99, 415)
+End Sub
+
+Public Function ModulusZero(ByVal dblInputValue As Single, ByVal dblDivisor As Single) As Boolean
+    'This procedure checks to see if any integer or decimal is equally divisible by the
+    'provided divisor. Note that the precision of the input value is evaluated at the same level
+    'as the precision of the divisor.
+    If dblInputValue = dblDivisor Then
+        'If the numbers are the same, then they are equally divisble.
+        ModulusZero = True
+    ElseIf dblInputValue < dblDivisor Then
+        'If the input value is smaller than the divisor, then it cannot be equally divisible.
+        ModulusZero = False
+    Else
+        'Microsoft warns about comparisons of doubles and singles and that they may not always
+        'evaluate how you exptect them to. Therefore, I treat their results of the maths as
+        'strings.
+        If InStr(CStr(dblInputValue), ".") = 0 And InStr(CStr(dblDivisor), ".") = 0 Then
+            ModulusZero = IIf(InStr(CStr(CLng(dblInputValue) / CLng(dblDivisor)), ".") > 0, False, True)
+        Else
+            ModulusZero = IIf(InStr(dblInputValue / dblDivisor, ".") > 0, False, True)
+        End If
+    End If
+End Function
+
+