The Solution is the Long Math module. In this module there are 4
functions: `IntAddition`

, `IntSubtraction`

, `IntMultiply`

and `IntDivide`

.
Each takes two strings, and each performs the the operation in the title on the strings, which are suppose to contain these massively large integers.

This is version 1 of this code. I am currently working on a second version that will take and handle positive and negative numbers, be more fault tolerent, and work with decimals. The attached source code download also includes a demonstration project.

`'******************************************************************`

' Extra Long Integer Mathematics

'

' Programmer: Eric L. Truitte

' Contact Info: etruitte@programmer.net

' Date Created: April 3, 2003

' Purpose:

' Functions to handle mathmatics too large for

' conventional variables and operations.

'

' Copyright notice:

' This code is subject to the GNU General Public License.

' If you make changes, add a Revision note.

' This code is Open Source and should remain so.

'

'******************************************************************

' Revision Date: April 5, 2003

' Programmer: Eric L. Truitte

' Details:

' Addition and Multiplication tested and finallized

' Update comments

'

' Revision Date: February 19, 2005

' Programmer: Eric L. Truitte

' Details:

' Finallized Subtraction and Division

' Updated comments

'

'******************************************************************

'Notes:

'

'THERE IS CURRENTLY VERY LITTLE FAULT TOLERANCE IN THE FUNCTIONS

'entering a non-numeric character will cause errors. If you intend to use these,

'make sure what you pass in is just a string of integers.

'

'To do list:

' Decimals

' Negatives

' Multi-string length numbers

'

'What is the largest number that can be handled at this point?

' A String can hold roughly the largest size of a Long value in characters.

' That is the number of potential digits you can have in a number.

'

'I will be working on a decimal handling, possitive/negative, and arrays of strings

'containing segments of an Extra Long Integer. I hope if you have need of such

'capacity of numbers that you have a system that can adequitely handle both

'system overhead and processor side-effects from overclocking for an extended

'period of time.

'

'******************************************************************

Public STOPNOW As Boolean 'This is a sentinel variable used in the demo form

Public Function IntAddition(ByVal FirstNum As String, ByVal SecondNum As String) As String

Dim a As Long, DifLen As Long, TempStr As String, TempNum As Integer

Dim Num1 As String, Num2 As String, TempNum1 As Integer, TempNum2 As Integer

Dim CarryOver As Integer, LeftOvers As Long

'Setup the numbers so that they are easier to handle.

'I originally had about 10 nested if statements that this block

'of code simplifies Dramatically.

If Len(FirstNum) >= Len(SecondNum) Then

Num1 = FirstNum

Num2 = SecondNum

Else

Num2 = FirstNum

Num1 = SecondNum

End If

'Just setup some of the variables that need an initial value

DifLen = Len(Num1) - Len(Num2)

CarryOver = 0

LeftOvers = DifLen

'Ok, now for the real math. Looping from the end of the numbers

'just like our preschool teachers taught us, we add numbers that

'line up in the 'places' (I.E. ones, tens, hundreds, thousands, etc)

For a = Len(Num2) To 1 Step -1

TempNum1 = Int(Mid(Num1, a + DifLen, 1))

TempNum2 = Int(Mid(Num2, a, 1))

TempNum = TempNum1 + TempNum2 + CarryOver

CarryOver = TempNum \ 10

TempStr = (TempNum - (CarryOver * 10)) & TempStr

DoEvents

If STOPNOW = True Then GoTo StopAdd

Next a

'What do we do if there is a 1 or a 2 that carries over outside the

'numbers that line up in the places, well, we do the following block of

'code. The do loop is used incase we get a situation like this:

'

' 199999 When you add 1 to a set of nines it continues to

' _+___1 Carry over until it hits the first digit

' 200000

Do Until CarryOver = 0 Or LeftOvers = 0

TempNum = Int(Mid(Num1, LeftOvers, 1)) + CarryOver

CarryOver = TempNum \ 10

TempStr = (TempNum - (CarryOver * 10)) & TempStr

LeftOvers = LeftOvers - 1

Loop

'Since there are two possible ways of exiting the Loop above, we need

'to test and apply the other variable and its associated values in the following

'two if statements.

'Handle a possible carryover that will drop off the front end creating a new place.

If CarryOver > 0 Then TempStr = CarryOver & TempStr

'add any of the numbers that are remaining on the left side of the longer string

If LeftOvers > 0 Then TempStr = Left(Num1, LeftOvers) & TempStr

'and return the value

StopAdd:

IntAddition = TrimZeros(TempStr)

End Function

Public Function IntMultiply(ByVal FirstNum As String, ByVal SecondNum As String) As String

Dim ZeroStr As String

Dim a As Long, b As Long, Multiplier1 As Integer, Multiplier2 As Integer

Dim Num As Integer, CarryOver As Integer, TempStr As String, TallyStr As String

'THIS FUNCTION IS COMPLETE AND WORKS

'This function can handle two extra longs. It cycles through

'the firstnum one digit at a time from secondnum.

'this function works on the distrubution Principle of Multiplication:

' 9999 * 222 = (9999 * 2) + (9999 * 20) + (9999 * 200)

'

'The zero's are concatinated on after the multiplication takes place.

'

'This function is dependent on the IntAddition function above.

For a = Len(FirstNum) To 1 Step -1

'setup variables for this loop of multiplication

TempStr = ""

CarryOver = 0

Multiplier1 = Mid(FirstNum, a, 1)

'Multiply one digit at a time from right to left

For b = Len(SecondNum) To 1 Step -1

Multiplier2 = Mid(SecondNum, b, 1)

Num = (Multiplier1 * Multiplier2) + CarryOver

CarryOver = Num \ 10

TempStr = (Num - (CarryOver * 10)) & TempStr

Next b

'Check to see if the multiplication added a new digit

If CarryOver > 0 Then TempStr = CarryOver & TempStr

'Add the zeros

TempStr = TempStr & ZeroStr

TallyStr = IntAddition(TempStr, TallyStr)

ZeroStr = ZeroStr & "0"

DoEvents

'sentinel

If STOPNOW = True Then GoTo StopMultiply

Next a

StopMultiply:

IntMultiply = TrimZeros(TallyStr)

End Function

Public Function TrimZeros(ByVal Num As String) As String

Dim a As Long, TempStr As String

For a = 1 To Len(Num)

If Mid(Num, a, 1) <> 0 Then GoTo YuckFu

Next a

TrimZeros = "0"

Exit Function

YuckFu:

TrimZeros = Mid(Num, a, Len(Num) - a + 1)

End Function

Public Function IntSubtract(ByVal FirstNum As String, ByVal SecondNum As String) As String

'***

'DO NOT change the integers to bytes, there are negative values in this function

'***

Dim Num1 As String, Num2 As String, a As Long, Neg As Boolean, DifLen As Long

Dim TempStr As String, TempNum1 As Integer, TempNum2 As Integer

Dim TempNum As Integer, Barrow As Byte

'This function operates on a theory known as Two-Compliment.

'If you want to know more, look for it at www.mathforum.com

'This function works great now

'This block of code arranges the numbers into the Num1 and Num2 based on

'which number is larger. This prevents a great number of errors if the numbers

'dont line up, or if the larger number is taken from the smaller number.

If Len(FirstNum) > Len(SecondNum) Then

Num1 = FirstNum

Num2 = SecondNum

Neg = False

ElseIf Len(FirstNum) < Len(SecondNum) Then

Num1 = SecondNum

Num2 = FirstNum

Neg = True

Else

'In the case that the strings are of equal length we have this pretty little

'set of code to find which number has the first larger digit.

For a = 1 To Len(FirstNum)

If Int(Mid(FirstNum, a, 1)) > Int(Mid(SecondNum, a, 1)) Then

Num1 = FirstNum

Num2 = SecondNum

Neg = False

GoTo ContinSubtraction

ElseIf Int(Mid(FirstNum, a, 1)) < Int(Mid(SecondNum, a, 1)) Then

Num1 = SecondNum

Num2 = FirstNum

Neg = True

GoTo ContinSubtraction

End If

DoEvents

'sentinel

If STOPNOW = True Then GoTo ExitFunction

Next a

'In the case that no larger digit is found, then guess what, its a perfect

'subtraction, so we don't need to do the function, just assign a 0 outside the end.

GoTo ExitFunction

End If

ContinSubtraction:

'If we have a difference in length then ajust with 0's that will not affect the calculations.

'This allows us to get all the digits into the final out number.

DifLen = Len(Num1) - Len(Num2)

Num2 = String(DifLen, "0") & Num2

Barrow = 0

'lets do some math

For a = Len(Num2) To 1 Step -1

'Pick out the individual digit from each number

TempNum1 = Int(Mid(Num1, a, 1)) - Barrow

TempNum2 = Int(Mid(Num2, a, 1))

Barrow = 0

'Perform single digit subraction using the Two Compliment theory

If TempNum1 >= TempNum2 Then

TempNum = TempNum1 - TempNum2

ElseIf TempNum1 < TempNum2 Then

TempNum = (TempNum1 + 10) - TempNum2

Barrow = 1

End If

'Assign new digit to the final string.

TempStr = CStr(TempNum) & TempStr

DoEvents

'sentinel

If STOPNOW = True Then GoTo ExitFunction

Next a

'now, since we are subtracting, we need to determine if the number being returned is a negative.

'Just to note, the Trim is to remove unneccsary zero's at the head(left) of the return number.

If Neg = True Then

IntSubtract = "-" & TrimZeros(Trim(TempStr))

Else

IntSubtract = TrimZeros(Trim(TempStr))

End If

Exit Function

ExitFunction:

IntSubtract = 0

End Function

Public Function IntDivide(ByVal FirstNum As String, ByVal SecondNum As String) As String

'Before we even alocate memory for variables, test for some very important error values

If Len(FirstNum) < Len(SecondNum) Or InStr(1, IntSubtract(FirstNum, SecondNum), "-") > 0 Then

MsgBox "Fault: Extra Long Division does not support dividing a shorter number by a longer number, as this requires decimals which are not currently handled."

ElseIf TrimZeros(SecondNum) = "" Then

MsgBox "Fault: Cannot divide by Zero."

ElseIf TrimZeros(SecondNum) = "" Then

GoTo EndFunc

Else

GoTo continDivide

End If

GoTo ExitDivide

'After passing the error checking, lets get started with some division

continDivide:

Dim Num1 As String, DivTotal As String, DivMult As String

Dim DifLen As Long, DivSub As String, TempNum As String

'Initiallize values

Num1 = FirstNum

DivTotal = "0"

DifLen = (Len(Num1) - Len(SecondNum))

DivMult = String(DifLen, "0")

DivSub = SecondNum & DivMult

'Lets do some division

Do Until (Len(Num1) < Len(SecondNum) Or Num1 = "0" Or (InStr(1, IntSubtract(Num1, SecondNum), "-") > 0 And DivMult = "")) Or STOPNOW = True

'The way this division works is it subtracts values from the divided number

'until no more can be subtracted. This sets up a the largest possible number

'that can be subtracted from the number so that you remove larger chucks of

'Numbers at a time and waste less CPU Cycles doing it.

If DifLen >= 0 Then DivMult = String(DifLen, "0")

DivSub = SecondNum & DivMult

If InStr(1, IntSubtract(Num1, DivSub), "-") > 0 Then

If DifLen > 0 Then

DivMult = String(DifLen - 1, "0")

DivSub = SecondNum & DivMult

Else

Exit Do

End If

End If

'Perform the accually math. DivTotal adds up how many times the original

'number has been subtracted from the divided number. Num1 is the working

'number.

DivTotal = IntAddition(DivTotal, "1" & DivMult)

Num1 = IntSubtract(Num1, DivSub)

DifLen = Len(Num1) - Len(SecondNum)

DoEvents

'sentinel

If STOPNOW = True Then GoTo ExitDivide

Loop

'Since there are no decimals, we return return the devide results with a remainder.

IntDivide = DivTotal & "r" & Num1

EndFunc:

Exit Function

ExitDivide:

IntDivide = "NaN"

End Function

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