That was funny. To instantiate Font in .NET 2.0 you call one of many Font's constructors. Basically you provide a font family (either an object of the class FontFamily or a name of one as a string), size and FontStyle. There is a constructor that doesn't take a FontStyle, it assumes the FontStyle.Regular as default.
A FontFamily is a collection of fonts. Arial family will have Arial Regular, Arial Bold, Arial Bold Italic...
Now, in theory there might be a family with only Bold Italic. The requirement is in case there is no style the user requested, give back a font from the family in any available style.
Unfortunatelly in FontFamily object there is no property returning available members of the family, but there is a method for checking if given FontStyle is available.
FontStyle is an enumeration with the following members.
FontStyle.Regular = 0 FontStyle.Bold = 1 FontStyle.Italic = 2 FontStyle.Underline = 4 FontStyle.Strikeout = 8
A font style can be created by bitwise or-ing the values. For example Bold Italic can be defined by FontStyle.Bold | FontStyle.Italic wihich makes 3.
How to generate all possible combinations?
for i in xrange(16):
fontStyle = Enum.ToObject(FontStyle, i)
if fontFamily.IsStyleAvailable(fontStyle):
return Font(fontFamily, fontSize, fontStyle)
Integers from 0 to 15 form all valid combinations of the FontStyle. This sixteen numbers can be represented with 4 bits, 15's binary representation having them all lit (8 + 4 + 2 + 1).
