Converting Numbers to Words Part III

See Converting Numbers to Words Part II

My tests work from 0-99. The next test will test numbers between 100-199.

Sub TEST_OneHundred()

Debug.Assert NumbersToWords(100) = "one hundred"
Debug.Assert NumbersToWords(110) = "one hundred ten"
Debug.Assert NumbersToWords(119) = "one hundred nineteen"
Debug.Assert NumbersToWords(120) = "one hundred twenty"
Debug.Assert NumbersToWords(121) = "one hundred twenty-one"
Debug.Assert NumbersToWords(150) = "one hundred fifty"
Debug.Assert NumbersToWords(188) = "one hundred eighty-eight"
Debug.Assert NumbersToWords(199) = "one hundred ninety-nine"

End Sub

A haphazard selection of numbers including the edge cases.

Function NumbersToWords(ByVal dNumbers As Double) As String

Dim vaSingles As Variant
Dim vaTens As Variant
Dim sReturn As String

vaSingles = Split("zero,one,two,three,four,five,six,seven,eight,nine,ten,eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen,nineteen", ",")
vaTens = Split("NA,NA,twenty,thirty,forty,fifty,sixty,seventy,eighty,ninety", ",")

If dNumbers >= 100 Then
sReturn = "one hundred"
If dNumbers Mod 100 <> 0 Then
If dNumbers - 100 > 19 Then
sReturn = sReturn & Space(1) & vaTens((dNumbers - 100) \ 10)
If (dNumbers - 100) Mod 10 <> 0 Then
sReturn = sReturn & "-" & vaSingles((dNumbers - 100) - (((dNumbers - 100) \ 10) * 10))
End If
Else
sReturn = sReturn & Space(1) & vaSingles(dNumbers - 100)
End If
End If

ElseIf dNumbers > 19 Then
sReturn = vaTens(dNumbers \ 10)
If dNumbers Mod 10 <> 0 Then
sReturn = sReturn & "-" & vaSingles(dNumbers - ((dNumbers \ 10) * 10))
End If
Else
sReturn = vaSingles(dNumbers)
End If

NumbersToWords = Trim$(sReturn)

End Function

And all tests pass. Back in the first post of this series I said that I hoped it would be obvious when I need to refactor. Well if this isn’t a frying pan to the face, I don’t know what is. Way too much repetition, for one. I need to introduce a “remainder” variable, so that once I process the hundred part, I can send the remainder to process the tens, and the remainder from that to the less than 19 part.

Function NumbersToWords(ByVal dNumbers As Double) As String

Dim vaSingles As Variant
Dim vaTens As Variant
Dim sReturn As String
Dim dRemainder As Double

vaSingles = Split("zero,one,two,three,four,five,six,seven,eight,nine,ten,eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen,nineteen", ",")
vaTens = Split("zero,zero,twenty,thirty,forty,fifty,sixty,seventy,eighty,ninety", ",")

dRemainder = dNumbers

If dRemainder >= 100 Then
sReturn = "one hundred" & Space(1)
dRemainder = dRemainder - (dRemainder \ 100) * 100
End If

If dRemainder > 19 Then
sReturn = sReturn & vaTens(dRemainder \ 10)
dRemainder = dRemainder - (dRemainder \ 10) * 10
End If

If dRemainder > 0 Then
If Right(sReturn, 1) = "y" Then
sReturn = sReturn & "-"
End If

sReturn = sReturn & vaSingles(dRemainder)
End If

NumbersToWords = Trim$(sReturn)

End Function

That looks much better, but it doesn’t pass the zero test. I don’t like special cases, but zero might just be one, so I’m going to force it. My conditional on whether to include a hyphen checks to see if the answer so far ends in “y”. That seems a little hokey, but it works. I could test for mod10 and set a Boolean variable in the If block above, but I’m not sure what I gain, so there it stays.

Refactoring in this way also makes the next bit of testing code painfully obvious. I’m hardcoding “one hundred”, but with vaSingles sitting right there, I don’t know why I can’t go above 199 pretty easily. So I’ll write that next test.

Sub TEST_Hundreds()

Debug.Assert NumbersToWords(200) = "two hundred"
Debug.Assert NumbersToWords(310) = "three hundred ten"
Debug.Assert NumbersToWords(419) = "four hundred nineteen"
Debug.Assert NumbersToWords(520) = "five hundred twenty"
Debug.Assert NumbersToWords(621) = "six hundred twenty-one"
Debug.Assert NumbersToWords(750) = "seven hundred fifty"
Debug.Assert NumbersToWords(888) = "eight hundred eighty-eight"
Debug.Assert NumbersToWords(999) = "nine hundred ninety-nine"

End Sub

Instead of hardcoding “one hundred”, I’ll pull the property number from vaSingles. This also shows my brute force zero fix.

Function NumbersToWords(ByVal dNumbers As Double) As String

Dim vaSingles As Variant
Dim vaTens As Variant
Dim sReturn As String
Dim dRemainder As Double

vaSingles = Split("zero,one,two,three,four,five,six,seven,eight,nine,ten,eleven,twelve,thirteen,fourteen,fifteen,sixteen,seventeen,eighteen,nineteen", ",")
vaTens = Split("zero,zero,twenty,thirty,forty,fifty,sixty,seventy,eighty,ninety", ",")

If dNumbers = 0 Then
sReturn = "zero"
Else

dRemainder = dNumbers

If dRemainder >= 100 Then
sReturn = sReturn & vaSingles(dRemainder \ 100) & " hundred "
dRemainder = dRemainder - (dRemainder \ 100) * 100
End If

If dRemainder > 19 Then
sReturn = sReturn & vaTens(dRemainder \ 10)
dRemainder = dRemainder - (dRemainder \ 10) * 10
End If

If dRemainder > 0 Then
If Right(sReturn, 1) = "y" Then
sReturn = sReturn & "-"
End If

sReturn = sReturn & vaSingles(dRemainder)
End If
End If

NumbersToWords = Trim$(sReturn)

End Function

All tests pass. And the code doesn’t look too bad. Only infinity numbers left to test. Here’s what my main testing procedure looks like now, as if you couldn’t guess.

Sub TEST_All()

TEST_Singles
TEST_Tens
TEST_OneHundred
TEST_Hundreds

Debug.Print "tests passed"

End Sub

Posted in VBA

2 thoughts on “Converting Numbers to Words Part III

  1. I remember posting a recursive function for this to StackOverflow a while ago:

    http://stackoverflow.com/a/5232125

    I think your code is more elegant, but you’ll run into issues with tens of thousands, and the hierarchical conventions for millions and billions. For example, you can express 2^50 as:

    “Eleven Hundred and Twenty-Five Trillion, Eight Hundred and Ninety-Nine Billion, Nine Hundred and Six Million, Eight Hundred and Forty-Two Thousand, Six Hundred and Twenty One Trillion, Ninety-Nine Billion, Five Hundred and Eleven Million, Six Hundred and Twenty-Seven Thousand, Seven Hundred and Seventy-Six”

    I do not anticipate that you will need to express this kind of number in text on a regular basis. However, there is some instructional value in demonstrating an application of recursion to beginners in programming.

    Public Function SayNumber(ByVal InputNumber As Double, _
    Optional DecimalPlaces As Integer = 0 _
    ) As String

    ' Return the integer portion of the number in formatted English words

    ' Return the fractional part of the number as 'point' and a series of
    ' single-numeral words, up to the precision specified by 'DecimalPlaces'

    ' SayNumber(17241021505)
    ' "Seventeen Billion, Two Hundred and Forty-One Million,
    ' Twenty-One Thousand, Five Hundred and Five"

    ' SayNumber(Sqr(2), 6)
    ' "One point Four One Four Two One Four"

    ' Note that nothing after the decimal point will be returned if InputNumber is an integer

    ' Nigel Heffernan, December 2008

    Application.Volatile False
    On Error Resume Next

    Dim arrDigits(0 To 9) As String
    Dim arrTeens(10 To 19) As String
    Dim arrTens(2 To 9) As String

    Dim i As Integer
    Dim i10 As Integer
    Dim i20 As Integer
    Dim dblRemainder As Double
    Dim dblMain As Double
    Dim iRemainder As Long
    Dim iMain As Long
    Dim dblFraction As Double
    Dim strFraction As String

    Dim strMinus As String
    Dim str1 As String
    Dim str2 As String
    Dim str3 As String

    If Application.EnableEvents = False Then
    Exit Function
    End If

    arrDigits(0) = "Zero"
    arrDigits(1) = "One"
    arrDigits(2) = "Two"
    arrDigits(3) = "Three"
    arrDigits(4) = "Four"
    arrDigits(5) = "Five"
    arrDigits(6) = "Six"
    arrDigits(7) = "Seven"
    arrDigits(8) = "Eight"
    arrDigits(9) = "Nine"

    arrTeens(10) = "Ten"
    arrTeens(11) = "Eleven"
    arrTeens(12) = "Twelve"
    arrTeens(13) = "Thirteen"
    arrTeens(14) = "Fourteen"
    arrTeens(15) = "Fifteen"
    arrTeens(16) = "Sixteen"
    arrTeens(17) = "Seventeen"
    arrTeens(18) = "Eighteen"
    arrTeens(19) = "Nineteen"

    arrTens(2) = "Twenty"
    arrTens(3) = "Thirty"
    arrTens(4) = "Forty"
    arrTens(5) = "Fifty"
    arrTens(6) = "Sixty"
    arrTens(7) = "Seventy"
    arrTens(8) = "Eighty"
    arrTens(9) = "Ninety"

    If InputNumber < 0 Then strMinus = "Minus " InputNumber = Abs(InputNumber) End If If DecimalPlaces < 1 Then strFraction = "" Else dblFraction = InputNumber - Fix(InputNumber) If dblFraction = 0 Then strFraction = "" Else strFraction = " point" str1 = Format(dblFraction, "0." & String(DecimalPlaces, "0")) For i = 1 To DecimalPlaces str2 = MID(str1, i + 2, 1) strFraction = strFraction & " " & arrDigits(CInt(str2)) str2 = "" Next i str1 = "" End If End If If InputNumber < 10 Then str1 = arrDigits(InputNumber) ElseIf InputNumber < 19 Then str1 = arrTeens(InputNumber) ElseIf InputNumber < 100 Then iMain = InputNumber \ 10 str1 = arrTens(iMain) iRemainder = InputNumber Mod 10 If iRemainder > 0 Then
    str2 = "-" & arrDigits(iRemainder)
    End If

    ElseIf InputNumber < 1000 Then iMain = InputNumber \ 100 str1 = arrDigits(iMain) & " Hundred" iRemainder = InputNumber Mod 100 If iRemainder > 0 Then
    str2 = " and " & SayNumber(iRemainder)
    End If

    ElseIf InputNumber < 2000 Then iMain = InputNumber \ 100 str1 = arrTeens(iMain) & " Hundred" iRemainder = InputNumber Mod 100 If iRemainder > 0 Then
    str2 = " and " & SayNumber(iRemainder)
    End If

    ElseIf InputNumber < 1000000 Then iMain = InputNumber \ 1000 str1 = SayNumber(iMain) & " Thousand" iRemainder = InputNumber Mod 1000 If iRemainder > 0 Then
    str2 = ", " & SayNumber(iRemainder)
    End If

    ElseIf InputNumber < (10 ^ 9) Then iMain = InputNumber \ (10 ^ 6) str1 = SayNumber(iMain) & " Million" iRemainder = InputNumber Mod (10 ^ 6) If iRemainder > 0 Then
    str2 = ", " & SayNumber(iRemainder)
    End If

    ElseIf InputNumber < (10 ^ 12) Then ' we'll hit the LongInt arithmetic operation limit at ~2.14 Billion str3 = Format(InputNumber, "0") dblMain = CDbl(Left(str3, Len(str3) - 9)) str1 = SayNumber(dblMain) & " Billion" dblRemainder = CDbl(Right(str3, 9)) If dblRemainder > 0 Then
    str2 = ", " & SayNumber(dblRemainder)
    End If

    ElseIf InputNumber < 1.79769313486231E+308 Then ' This will generate a recursive string of 'Trillions' str3 = Format(InputNumber, "0") dblMain = CDbl(Left(str3, Len(str3) - 12)) str1 = SayNumber(dblMain) & " Trillion" dblRemainder = CDbl(Right(str3, 12)) If dblRemainder > 0 Then
    str2 = ", " & SayNumber(dblRemainder)
    End If

    Else ' exceeds the specification for double-precision floating-point variables

    str1 = "#Overflow."

    End If

    SayNumber = strMinus & str1 & str2 & strFraction

    End Function

    I will leave the corresponding code in French as an exercise for the reader.

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