Library tutorials & articles
Number Systems
By Mitch Dusina, published on 16 Dec 2005
Page 3 of 5
- The Different Systems
- Any To Decimal
- Decimal To Any
- Any To Any
- Roman Numerals
Decimal To Any
Converting from the Decimal system to others invloves some slightly
complex math. To fully understand this code you will need to
understand how
Mod
functions.
Public Function DecimalToAny(ByVal Number As Long, ByVal Base As Short)_
As String
Dim strNumberOutput As String
'These bounds are in place because they will hold the Long's
'largest possible value.
'An Int16 and String array are both used for readability.
'You only need to use a string array
Dim intRemander(36) As Int16
Dim strRemander(36) As String
Dim IsNegative As Boolean
Dim i As Int16
'Roman Numerals
If Base = 0 Then
Return ToRomanNumeral(Number)
End If
'Supported Bases (1-9 A-Z)
If Base > 36 Or Base < 0 Then Throw New_
Exception("Start Base can only be 0 through 36_
(0-9, A-Z, or Roman Numerals)")
'Return 0 unless Unary (there is no number 0 in unary)
If Number = 0 And Base <> 1 Then
Return "0"
ElseIf Number = 0 And Base = 1 Then
Return ""
End If
'If needed, Adds the Negative sign after computation
If Number < 0 Then
IsNegative = True
Number *= -1
End If
'Special circumstances for Unary
If Base = 1 Then
Dim strUnary As String = ""
If IsNegative = True Then
strUnary = "-"
End If
'Think tick marks
For i = 0 To Number - 1
strUnary &= "1"
Next
Return strUnary
Exit Function
End If
'Count the remainders
Do Until Number = 0
intRemander(i) = Number Mod Base
strRemander(i) = intRemander(i).ToString
Number -= Number Mod Base
Number = (Number / Base)
i += 1
Loop
'Take off unused elements
ReDim Preserve intRemander(i - 1)
ReDim Preserve strRemander(i - 1)
'Converts > 9 to Text (10 = A)
For i = 0 To intRemander.GetUpperBound(0)
If Int(strRemander(i)) > 9 Then
strRemander(i) = Chr(strRemander(i) + 55)
End If
Next
'Concatinate the Remanders backwards
For i = intRemander.GetUpperBound(0) To 0 Step -1
strNumberOutput &= strRemander(i)
Next
'If needed, add that negative sign
If IsNegative = True Then strNumberOutput = "-" & strNumberOutput
Return strNumberOutput
End Function
Languages I know (from most comfortable with, to least comfortable with): VB.NET, C#, VB6, and C/C++
I'm 19 years old and pursuing a MS in Computer Science at the Colorado School of Mines. M...
Related articles
Related discussion
-
Run-time error '91'
by converter2009 (1 replies)
-
VB6 Runtime error 381 subsript out of range Error
by Uncle (2 replies)
-
passing and reading parameters from using Shell
by jigartoliya (0 replies)
-
Convert C++ code to VB6
by mawcot (4 replies)
-
listbox scrollbar
by Dennijr (10 replies)
Related podcasts
-
Christian Beauclair
14 mai 2008 (�mission #0074) ::.Christian Beauclair: Stratégies de migration VB6 vers .NET Nous discutons avec Christian Beauclair des stratégies de migration VB6 vers .NET. Entre autres, nous discutons comment utiliser le "VB 6 Code Advisor" et le "Interop Forms Toolkit" pour ajouter la puiss...
It looks good, the only thing I can spot is the line where you catch the Exception. Not finding an extended numeral is expected (and common at that), so an Exception shouldn't be used to control the flow of code.
Although I can't think of an easy way around it since they removed Enum.TryParse (2.0 and later I think).
private int ParseRomanNumeral(string numeral)
{
numeral = numeral.ToUpper();
char[] characters = numeral.ToCharArray();
int[] parseBuf = new int[numeral.Length];
int result = 0;
bool bSkip = false;
for (int i = numeral.Length - 1; i >= 0; i--)
{
if (bSkip)
{
bSkip = false; // Skip this digit because it is part of a 2-digit numeral
continue;
}
parseBuf[i] = (int)Enum.Parse(typeof(RomanNumerals), characters[i].ToString());
if (parseBuf[i] % 5 == 0 && i > 0)
{
string extended = characters[i-1].ToString() + characters[i].ToString();
try
{
parseBuf[i] = (int)Enum.Parse(typeof(RomanNumerals), extended);
bSkip = true;
}
catch(ArgumentException)
{
// Ignore this exception. It just means we didn't find an extended numeral
}
}
}
for (int i = 0; i < numeral.Length; i++)
{
result += parseBuf[i];
}
return result;
}
private string ConvertToRomanNumeral(int num)
{
string sResult = "";
sResult = Enum.GetName(typeof(RomanNumerals), num);
if (sResult == null)
{
string[] names = Enum.GetNames(typeof(RomanNumerals));
int[] vals = (int[])Enum.GetValues(typeof(RomanNumerals));
for (int i = names.Length - 1; i > 0; i--) // Don't process the 0th element, since it is 0 and results in bad math
{
int temp = num / vals[i];
num -= temp * vals[i];
for (int j = 0; j < temp; j++)
{
sResult += names[i];
}
}
}
return sResult;
}
This thread is for discussions of Number Systems.