C# – What does “Method …ClassInitialize has wrong signature …” mean

The [Flags] attribute should be used whenever the enumerable represents a collection of possible values, rather than a single value. Such collections are often used with bitwise operators, for example:

var allowedColors = MyColor.Red | MyColor.Green | MyColor.Blue;

Note that the [Flags] attribute doesn't enable this by itself - all it does is allow a nice representation by the .ToString() method:

enum Suits { Spades = 1, Clubs = 2, Diamonds = 4, Hearts = 8 }
[Flags] enum SuitsFlags { Spades = 1, Clubs = 2, Diamonds = 4, Hearts = 8 }

...

var str1 = (Suits.Spades | Suits.Diamonds).ToString();
           // "5"
var str2 = (SuitsFlags.Spades | SuitsFlags.Diamonds).ToString();
           // "Spades, Diamonds"

It is also important to note that [Flags] does not automatically make the enum values powers of two. If you omit the numeric values, the enum will not work as one might expect in bitwise operations, because by default the values start with 0 and increment.

Incorrect declaration:

[Flags]
public enum MyColors
{
    Yellow,  // 0
    Green,   // 1
    Red,     // 2
    Blue     // 3
}

The values, if declared this way, will be Yellow = 0, Green = 1, Red = 2, Blue = 3. This will render it useless as flags.

Here's an example of a correct declaration:

[Flags]
public enum MyColors
{
    Yellow = 1,
    Green = 2,
    Red = 4,
    Blue = 8
}

To retrieve the distinct values in your property, one can do this:

if (myProperties.AllowedColors.HasFlag(MyColor.Yellow))
{
    // Yellow is allowed...
}

or prior to .NET 4:

if((myProperties.AllowedColors & MyColor.Yellow) == MyColor.Yellow)
{
    // Yellow is allowed...
}

if((myProperties.AllowedColors & MyColor.Green) == MyColor.Green)
{
    // Green is allowed...
}    

Under the covers

This works because you used powers of two in your enumeration. Under the covers, your enumeration values look like this in binary ones and zeros:

 Yellow: 00000001
 Green:  00000010
 Red:    00000100
 Blue:   00001000

Similarly, after you've set your property AllowedColors to Red, Green and Blue using the binary bitwise OR | operator, AllowedColors looks like this:

myProperties.AllowedColors: 00001110

So when you retrieve the value you are actually performing bitwise AND & on the values:

myProperties.AllowedColors: 00001110
             MyColor.Green: 00000010
             -----------------------
                            00000010 // Hey, this is the same as MyColor.Green!

The None = 0 value

And regarding the use of 0 in your enumeration, quoting from MSDN:

[Flags]
public enum MyColors
{
    None = 0,
    ....
}

Use None as the name of the flag enumerated constant whose value is zero. You cannot use the None enumerated constant in a bitwise AND operation to test for a flag because the result is always zero. However, you can perform a logical, not a bitwise, comparison between the numeric value and the None enumerated constant to determine whether any bits in the numeric value are set.

You can find more info about the flags attribute and its usage at msdn and designing flags at msdn

It's the null coalescing operator, and quite like the ternary (immediate-if) operator. See also ?? Operator - MSDN.

FormsAuth = formsAuth ?? new FormsAuthenticationWrapper();

expands to:

FormsAuth = formsAuth != null ? formsAuth : new FormsAuthenticationWrapper();

which further expands to:

if(formsAuth != null)
    FormsAuth = formsAuth;
else
    FormsAuth = new FormsAuthenticationWrapper();

In English, it means "If whatever is to the left is not null, use that, otherwise use what's to the right."

Note that you can use any number of these in sequence. The following statement will assign the first non-null Answer# to Answer (if all Answers are null then the Answer is null):

string Answer = Answer1 ?? Answer2 ?? Answer3 ?? Answer4;

Also it's worth mentioning while the expansion above is conceptually equivalent, the result of each expression is only evaluated once. This is important if for example an expression is a method call with side effects. (Credit to @Joey for pointing this out.)