What is the c# equivalent of Java DecimalFormat?

peter.murray.rust picture peter.murray.rust · Oct 16, 2009 · Viewed 10k times · Source

How would I convert the following code to C#

DecimalFormat form 
String pattern = "";
for (int i = 0; i < nPlaces - nDec - 2; i++) {
        pattern += "#";
}
pattern += "0.";
for (int i = nPlaces - nDec; i < nPlaces; i++) {
        pattern += "0";
}
form = (DecimalFormat) NumberFormat.getInstance();
DecimalFormatSymbols symbols = form.getDecimalFormatSymbols();
symbols.setDecimalSeparator('.');
form.setDecimalFormatSymbols(symbols);
form.setMaximumIntegerDigits(nPlaces - nDec - 1);
form.applyPattern(pattern);

EDIT The particular problem is that I do not wish the decimal separator to change with Locale (e.g. some Locales would use ',').

Answer

manji picture manji · Oct 16, 2009

For decimal separator you can set it in a NumberFormatInfo instance and use it with ToString:

    NumberFormatInfo nfi = new NumberFormatInfo();
    nfi.NumberDecimalSeparator = ".";

    //** test **
    NumberFormatInfo nfi = new NumberFormatInfo();
    decimal d = 125501.0235648m;
    nfi.NumberDecimalSeparator = "?";
    s = d.ToString(nfi); //--> 125501?0235648

to have the result of your java version, use the ToString() function version with Custom Numeric Format Strings (i.e.: what you called pattern):

s = d.ToString("# ### ##0.0000", nfi);// 1245124587.23     --> 1245 124 587?2300
                                      //      24587.235215 -->       24 587?2352

System.Globalization.NumberFormatInfo