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.

--
Regards,
Tom Ogilvy


"DMB" wrote in message
...
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.

In article ,
DMB wrote:

this formula is giving me an error. I need the remainder. Am I doing this
wrong?

26 Mod 3

Try =MOD(26,3)

"DMB" wrote:

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

"DMB" wrote in message
...
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.
--
Regards,
Tom Ogilvy

"DMB" wrote in message
...
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