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MS Office Forum / Excel / Programming / January 2006Mod function not working right?
this formula is giving me an error. I need the remainder. Am I doing this wrong? 26 Mod 3 from the immediate window: ? 26 Mod 3 2 seems to work fine.  SignatureRegards, Tom Ogilvy > this formula is giving me an error. I need the remainder. Am I doing this > wrong? > > 26 Mod 3 Hard to know if you're doing it wrong, since you don't say what you're trying to do. Mod is an operator, so it only makes sense in an assignment statement. For instance: a = 26 Mod 3 returns 2 to the variable a. A bare 26 Mod 3 doesn't make sense - it's like putting the number 2 as a statement. > this formula is giving me an error. I need the remainder. Am I doing this > wrong? > > 26 Mod 3 Try =MOD(26,3) > this formula is giving me an error. I need the remainder. Am I doing this > wrong? > > 26 Mod 3 What answer are you expecting? Tim > this formula is giving me an error. I need the remainder. Am I doing this > wrong? > > 26 Mod 3 Function LumberLabel(width As Range, height As Range, material As Range, spacing As Range) As String Dim myWidth As String Dim myHeight As String Dim myMaterial As String Dim mySpacing As String myWidth = width myHeight = height myMaterial = material mySpacing = spacing 'Material type: DF #2, DF#1, PSL, ect. 'Add spacing ie @ 16" O.C. , @ 25" O.C. If mySpacing <> 0 And mySpacing <> "" Then mySpacing = " @ " & mySpacing & " in oc" 'Lumber sizing Dim x As Integer 'x = width Mod 0.5 Dim aWidth As Double aWidth = Round(myWidth) If ((aWidth / 2) - (aWidth \ 2)) <> 0 Then LumberLabel = "1. Will not work (myWidth / 2) = " & (myWidth / 2) & " and myWidth \ 2 = " & (myWidth \ 2) 'ElseIf x <> 0 Then LumberLabel = "2. x <> 0 = " & width Mod 0.5 Else LumberLabel = myWidth & " x " & myHeight & myMaterial & mySpacing End If 'This is giving me the error 'width Mod 0.5 'and this LumberLabel = "2. = " & 19.6 Mod 3.2 'A = 19 Mod 6.7 End Function ? 19.6 Mod 3.2 2 the equivalent of 20 mod 3 which is the expected behavior as stated in help on the MOD function. So you still didn't answer what you expect? width mod 0.5 gives me a divide by zero, so 0.5 is rounded down to 0 I suspect. (it is probably stored as .49999999999) I don't think the mod operator will work for you. SignatureRegards, Tom Ogilvy > Function LumberLabel(width As Range, height As Range, material As Range, > spacing As Range) As String [quoted text clipped - 36 lines] > > End Function |
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