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2205

answers:

5

Is there an efficient method of converting an integer into the written numbers for example:

String Written = IntegerToWritten(21);

would return "Twenty One"

Is there any way of doing this that doesn't involve a massive lookup table?

+1  A: 

why massive lookup table?

string GetWrittenInteger(int n)
{
  string[] a = new string[] {"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" }
  string[] b = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }
  string[] c = new string[] {"Twenty", "Thirty", "Forty", "Sixty", "Seventy", "Eighty", "Ninety"};
  string[] d = new string[] {"Hundred", "Thousand", "Million"}

  string s = n.ToString();
  for (int i = 0; i < s.Length; i++)
  {
    // logic (too lazy but you get the idea)
  }
}
lubos hasko
+5  A: 
Calanus
A: 

Here are the correct links to the posts by Justin Rogers: initial post, code only.

There is also an article on CodeProject that addresses this issue.

Yaakov Ellis
+2  A: 

I use this code.It is VB code but you can easily translate it to C#. It works

Function NumberToText(ByVal n As Integer) As String

Select Case n
Case 0
Return ""

Case 1 To 19
Dim arr() As String = {"One","Two","Three","Four","Five","Six","Seven", _
"Eight","Nine","Ten","Eleven","Twelve","Thirteen","Fourteen", _
"Fifteen","Sixteen","Seventeen","Eighteen","Nineteen"}
Return arr(n-1) & " "

Case 20 to 99
Dim arr() as String = {"Twenty","Thirty","Forty","Fifty","Sixty","Seventy","Eighty","Ninety"}
Return arr(n\10 -2) & " " & NumberToText(n Mod 10)

Case 100 to 199
Return "One Hundred " & NumberToText(n Mod 100)

Case 200 to 999
Return NumberToText(n\100) & "Hundreds " & NumberToText(n mod 100)

Case 1000 to 1999
Return "One Thousand " & NumberToText(n Mod 1000)

Case 2000 to 999999
Return NumberToText(n\1000) & "Thousands " & NumberToText(n Mod 1000)

Case 1000000 to 1999999
Return "One Million " & NumberToText(n Mod 1000000)

Case 1000000 to 999999999
Return NumberToText(n\1000000) & "Millions " & NumberToText(n Mod 1000000)

Case 1000000000 to 1999999999
Return "One Billion " & NumberTotext(n Mod 1000000000)

Case Else
Return NumberToText(n\1000000000) & "Billion " _
& NumberToText(n mod 1000000000)
End Select
End Function
Nick Masao
+22  A: 

This should work reasonably well:

public static class HumanFriendlyInteger
{
static string[] ones = new string[] { "", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine" };
static string[] teens = new string[] { "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" };
static string[] tens = new string[] { "Twenty", "Thirty", "Forty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" };
static string[] thousandsGroups = { "", " Thousand", " Million", " Billion" };

private static string FriendlyInteger(int n, string leftDigits, int thousands)
{
if (n == 0)
{
return leftDigits;
}
string friendlyInt = leftDigits;
if (friendlyInt.Length > 0)
{
friendlyInt += " ";
}

if (n < 10)
{
friendlyInt += ones[n];
}
else if (n < 20)
{
friendlyInt += teens[n - 10];
}
else if (n < 100)
{
friendlyInt += FriendlyInteger(n % 10, tens[n / 10 - 2], 0);
}
else if (n < 1000)
{
friendlyInt += FriendlyInteger(n % 100, (ones[n / 100] + " Hundred"), 0);
}
else
{
friendlyInt += FriendlyInteger(n % 1000, FriendlyInteger(n / 1000, "", thousands+1), 0);
}

return friendlyInt + thousandsGroups[thousands];
}

public static string IntegerToWritten(int n)
{
if (n == 0)
{
return "Zero";
}
else if (n < 0)
{
return "Negative " + IntegerToWritten(-n);
}

return FriendlyInteger(n, "", 0);
}

}

(Edited to make it considerably more concise.)

Wedge
I want to see the internationalised version.
John Nolan
It's interesting to note how many tiny differences there are between the above (US English) and a UK English equivalent, let alone other languages... :-)
Christian Hayter