What does the [Flags] Enum Attribute mean in C#?
Asked Answered
I

14

1753

From time to time I see an enum like the following:

[Flags]
public enum Options 
{
    None    = 0,
    Option1 = 1,
    Option2 = 2,
    Option3 = 4,
    Option4 = 8
}

I don't understand what exactly the [Flags] attribute does.

Anyone have a good explanation or example they could post?

Impede answered 12/8, 2008 at 4:9 Comment(4)
It's also worth noting, in addition to the accepted answer, that VB.NET actually requires [Flags] - at least according to the .NET guys: social.msdn.microsoft.com/forums/en-US/csharplanguage/thread/…Aliment
Note, not required in VB these days. Save behaviour as C# - just changes the ToString() output. Note, you can also do logical OR, WITHIN the Enum itself. Very cool. Cat = 1, Dog = 2, CatAndDog = Cat || Dog.Almshouse
@Almshouse You mean CatAndDog = Cat | Dog (the logical OR instead of the Conditional), I assume?Heroin
@DdW, partially correct: | should be used, but | is called the binary OR. II is the logical OR (that allows short-circuiting): At least according to Microsoft ;) msdn.microsoft.com/en-us/library/f355wky8.aspxAarhus
S
2531

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

Smite answered 12/8, 2008 at 5:10 Comment(14)
Flags itself does nothing. Also, C# does not require Flags per se. But the ToString implementation of your enum uses Flags, and so does Enum.IsDefined, Enum.Parse, etc. Try to remove Flags and look at the result of MyColor.Yellow | MyColor.Red; without it you get "5", with Flags you get "Yellow, Red". Some other parts of the framework also use [Flags] (e.g., XML Serialization).Baro
The Flags attribute has more of an effect than the ToString method. Visual studio uses it at least to some degree when using an Enum in the designer. See my question here: https://mcmap.net/q/22277/-using-a-or-39-ed-enum-in-a-custom-uitypeeditor/1111886Eveland
I prefer to use constants of the form A = 1 << 0, B = 1 << 1, C = 1 << 2 ... Much easier to read, understand, visually check and change.Chapen
.HasFlag(Enum flag) method takes parameter of Enum type, so the value type should be boxed which is bad for performanceScandura
@AlekseiChepovoi Enum derives from the ValueType class, this means that it is treated as a struct the same way as int, float, etc.Kezer
@borrrden, when we pass Enum class (class that defines HasFlag method), it is passed by reference, as I got it. I've read this in Jeffrey Richter's CLR via C# third edition, Enums and Bit Flags chapter.Scandura
@AlekseiChepovoi I also have that book, so I will check it. It seems to fly in the face of everything I know about C# though. Every article I can find says that enums are value objects that are not allocated on the heap.Kezer
@borrrden, hell yeahh! I found this: msdn.microsoft.com/en-us/library/system.enum.aspx - see "Remarks" part: "Enum is the base class for all enumerations in the .NET Framework." and "The enumeration does not explicitly inherit from Enum; the inheritance relationship is handled implicitly by the compiler." So, when you write: public enum bla bla bla - this is a value type. But, the HasFlag method wants you to give him an instance of System.Enum which is a class (reference type:)Scandura
Enum.IsDefined does not take the FlagsAttribute into account. None of the following return true, even with the attribute: Yellow | Green, "Yellow, Green", 3President
your example would make more sense from a Flag perceptive if blue = 2 and green = 3 as after all green is the combination of yellow and bluePhyllys
If you want to exclude a flag from the enumeration, use xor, which is ^ in C#. So if you have myProperties.AllowedColors = MyColor.Red | MyColor.Green | MyColor.Blue;. You can do: myProperties.AllowedColors = myProperties.AllowedColors ^ MyColor.Blue //myProperties.AllowedColors == MyColor.Red | Mycolor.GreenTareyn
I was looking what 10 represents... after some testing this corresponds to value 2 and value 8 or Green | Blue. also the values of Yellow = 1, Green = 2,... 8 helps to easily understand what a 9 (8+1) or 11 (8+2+1) means vs the shift representation (<<).Showboat
@JoshNoe The issue with that is that the XOR will flip the bit on if it is not already flipped on, so if you had myProperties.AllowedColors = MyColor.Red | MyColor.Green; and you attempted to exclude blue with myProperties.AllowedColors = myProperties.AllowedColors ^ MyColor.Blue;, you would end up including it rather than excluding it. Instead, you can use myProperties.AllowedColor &= int.MaxValue ^ MyColor.Blue; to turn off a bit without the risk of accidentally turning it on if it's already off.Loquacity
Where can i find the internal implementation of this? how the | operator return the Enum type? I'm trying to find the implementation in Java, Scala and I cannot find any reference to such thing. how its internal implemented?Perrie
M
892

You can also do this

[Flags]
public enum MyEnum
{
    None   = 0,
    First  = 1 << 0,
    Second = 1 << 1,
    Third  = 1 << 2,
    Fourth = 1 << 3
}

I find the bit-shifting easier than typing 4,8,16,32 and so on. It has no impact on your code because it's all done at compile time

Margetts answered 12/8, 2008 at 4:37 Comment(6)
"it's all done at compile time" this type of comment may be interesting to some, but I find it frustrating because it implies that say, 20 bit-shifts at run-time will have any measurable impact on performance. CPUs are very good at bit-shifts, enums are not going to have an arbitrary number of members, and this won't work for more than 32 or 64 elements anyway. Not remotely important as a performance issue.Siloa
@Siloa I agree, it's silly to be overly worried about the runtime performance of this, but all the same, I do think its nice to know that this isn't going to be inserting bitshifts anywhere you use the enum. More of a 'that's neat' thing rather than anything related to performanceMargetts
@JeremyWeir - Several bits are going to be set in a flag enumeration value. So your method of data analysis is what is improper. Run a procedure to Represent your integer value in binary. 131,072 [D] = 0000 0000 0000 0001 0000 0000 0000 0000 [B] [32].. With the 17th bit set, an enum value assigned 1<<17 is easily determinable. 0110 0011 0011 0101 0101 0011 0101 1011 [b] [32] .. the enum values assign 1<<31, 1<<30, 1<<26, 1<<25.. etc. etc. is determinable without even the aid of a calculator at all.. which I doubt you would be able to determine at all without getting the binary rep.Mausoleum
Also pointing out you can bit-shift from "previous" enum values rather than directly from numbers, i.e. Third = Second << 1 rather than Third = 1 << 2 -- see more complete description belowTurcotte
I like this, but once you get to the upper limits of the underlying enum type, the compiler doesn't warn against bit shifts such as 1 << 31 == -2147483648, 1 << 32 == 1, 1 << 33 == 2, and so on. By contrast, if you say ThirtySecond = 2147483648 for an int type enum, the compiler throws an error.Wicket
Just to add another option that's new with C# 7, you can use 0b1, 0b10 etc. (or 0b00000001 if you like everything lining up properly!)Thimbu
T
141

Combining answers https://mcmap.net/q/22276/-what-does-the-flags-enum-attribute-mean-in-c (declaration via bit-shifting) and https://mcmap.net/q/22276/-what-does-the-flags-enum-attribute-mean-in-c (using combinations in declaration) you can bit-shift previous values rather than using numbers. Not necessarily recommending it, but just pointing out you can.

Rather than:

[Flags]
public enum Options : byte
{
    None    = 0,
    One     = 1 << 0,   // 1
    Two     = 1 << 1,   // 2
    Three   = 1 << 2,   // 4
    Four    = 1 << 3,   // 8

    // combinations
    OneAndTwo = One | Two,
    OneTwoAndThree = One | Two | Three,
}

You can declare

[Flags]
public enum Options : byte
{
    None    = 0,
    One     = 1 << 0,       // 1
    // now that value 1 is available, start shifting from there
    Two     = One << 1,     // 2
    Three   = Two << 1,     // 4
    Four    = Three << 1,   // 8

    // same combinations
    OneAndTwo = One | Two,
    OneTwoAndThree = One | Two | Three,
}

Confirming with LinqPad:

foreach(var e in Enum.GetValues(typeof(Options))) {
    string.Format("{0} = {1}", e.ToString(), (byte)e).Dump();
}

Results in:

None = 0
One = 1
Two = 2
OneAndTwo = 3
Three = 4
OneTwoAndThree = 7
Four = 8
Turcotte answered 16/5, 2013 at 15:3 Comment(2)
The combinations are a good recommendation, but I think the chained bit-shift would be prone to copy-and-paste errors such as Two = One << 1, Three = One << 1, etc... The incrementing integers of the form 1 << n are safer and the intent is clearer.Avowed
@RupertRawnsley to quote my answer: > Not necessarily recommending it, but just pointing out you canTurcotte
T
65

In extension to the accepted answer, in C#7 the enum flags can be written using binary literals:

[Flags]
public enum MyColors
{
    None   = 0b0000,
    Yellow = 0b0001,
    Green  = 0b0010,
    Red    = 0b0100,
    Blue   = 0b1000
}

I think this representation makes it clear how the flags work under the covers.

Thanet answered 20/7, 2017 at 11:9 Comment(1)
And in C# 7.2 it's even clearer with leading separator! 0b_0100Deserving
C
53

Please see the following for an example which shows the declaration and potential usage:

namespace Flags
{
    class Program
    {
        [Flags]
        public enum MyFlags : short
        {
            Foo = 0x1,
            Bar = 0x2,
            Baz = 0x4
        }

        static void Main(string[] args)
        {
            MyFlags fooBar = MyFlags.Foo | MyFlags.Bar;

            if ((fooBar & MyFlags.Foo) == MyFlags.Foo)
            {
                Console.WriteLine("Item has Foo flag set");
            }
        }
    }
}
Casuistry answered 12/8, 2008 at 4:32 Comment(1)
This example works even if you leave out [Flags]. It's the [Flags] part I'm trying to learn about.Gemot
C
46

When working with flags I often declare additional None and All items. These are helpful to check whether all flags are set or no flag is set.

[Flags] 
enum SuitsFlags { 

    None =     0,

    Spades =   1 << 0, 
    Clubs =    1 << 1, 
    Diamonds = 1 << 2, 
    Hearts =   1 << 3,

    All =      ~(~0 << 4)

}

Usage:

Spades | Clubs | Diamonds | Hearts == All  // true
Spades & Clubs == None  // true

 
Update 2019-10:

Since C# 7.0 you can use binary literals, which are probably more intuitive to read:

[Flags] 
enum SuitsFlags { 

    None =     0b0000,

    Spades =   0b0001, 
    Clubs =    0b0010, 
    Diamonds = 0b0100, 
    Hearts =   0b1000,

    All =      0b1111

}
Casanova answered 23/10, 2018 at 8:43 Comment(1)
Yeah, combining all answers in this page, I'm using regularly this one now, but with the leading separator 0b_1111Callboy
R
45

I asked recently about something similar.

If you use flags you can add an extension method to enums to make checking the contained flags easier (see post for detail)

This allows you to do:

[Flags]
public enum PossibleOptions : byte
{
    None = 0,
    OptionOne = 1,
    OptionTwo = 2,
    OptionThree = 4,
    OptionFour = 8,

    //combinations can be in the enum too
    OptionOneAndTwo = OptionOne | OptionTwo,
    OptionOneTwoAndThree = OptionOne | OptionTwo | OptionThree,
    ...
}

Then you can do:

PossibleOptions opt = PossibleOptions.OptionOneTwoAndThree 

if( opt.IsSet( PossibleOptions.OptionOne ) ) {
    //optionOne is one of those set
}

I find this easier to read than the most ways of checking the included flags.

Rebuke answered 12/8, 2008 at 18:40 Comment(5)
IsSet is an extension method I assume?Brilliant
Yeah - read the other question that I link to for details: stackoverflow.com/questions/7244Rebuke
.NET 4 adds a HasFlag method to enumerations, so you can do opt.HasFlag( PossibleOptions.OptionOne ) without having to write your own extensionsMargetts
Note that HasFlag is much slower than doing bitwise operations.Hak
@WaiHaLee You can use CodeJam.EnumHelper.IsFlagSet extension method which is compiled to a fast version using Expression: github.com/rsdn/CodeJam/wiki/M_CodeJam_EnumHelper_IsFlagSet__1Dinerman
S
26

@Nidonocu

To add another flag to an existing set of values, use the OR assignment operator.

Mode = Mode.Read;
//Add Mode.Write
Mode |= Mode.Write;
Assert.True(((Mode & Mode.Write) == Mode.Write)
  && ((Mode & Mode.Read) == Mode.Read)));
Schaffner answered 12/8, 2008 at 15:37 Comment(0)
T
19

To add Mode.Write:

Mode = Mode | Mode.Write;
Tho answered 12/8, 2008 at 5:59 Comment(1)
Or, for a shorter syntax: Mode |= Mode.Write;Footrope
C
16

Flags allow you to use bitmasking inside your enumeration. This allows you to combine enumeration values, while retaining which ones are specified.

[Flags]
public enum DashboardItemPresentationProperties : long
{
    None = 0,
    HideCollapse = 1,
    HideDelete = 2,
    HideEdit = 4,
    HideOpenInNewWindow = 8,
    HideResetSource = 16,
    HideMenu = 32
}
Childhood answered 12/8, 2008 at 4:24 Comment(1)
This is incorrect, you can use bitmasking even if enum is not marked as Flags.Erudite
R
16

There's something overly verbose to me about the if ((x & y) == y)... construct, especially if x AND y are both compound sets of flags and you only want to know if there's any overlap.

In this case, all you really need to know is if there's a non-zero value[1] after you've bitmasked.

[1] See Jaime's comment. If we were authentically bitmasking, we'd only need to check that the result was positive. But since enums can be negative, even, strangely, when combined with the [Flags] attribute, it's defensive to code for != 0 rather than > 0.

Building off of @andnil's setup...

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace BitFlagPlay
{
    class Program
    {
        [Flags]
        public enum MyColor
        {
            Yellow = 0x01,
            Green = 0x02,
            Red = 0x04,
            Blue = 0x08
        }

        static void Main(string[] args)
        {
            var myColor = MyColor.Yellow | MyColor.Blue;
            var acceptableColors = MyColor.Yellow | MyColor.Red;

            Console.WriteLine((myColor & MyColor.Blue) != 0);     // True
            Console.WriteLine((myColor & MyColor.Red) != 0);      // False                
            Console.WriteLine((myColor & acceptableColors) != 0); // True
            // ... though only Yellow is shared.

            Console.WriteLine((myColor & MyColor.Green) != 0);    // Wait a minute... ;^D

            Console.Read();
        }
    }
}
Rodmann answered 1/5, 2013 at 23:57 Comment(2)
Enum may be based on a signed type, so you should use "!= 0" instead of ">0".Maladapted
@JaimePardos -- As long as we keep them honest bytes, as I do in this example, there's no concept of negative. Just 0 to 255. As MSDN warns, "Use caution if you define a negative number as a flag enumerated constant because... [that] might make your code confusing and encourage coding errors." It's strange to think in terms of "negative bitflags"! ;^) I'll edit more in a bit. But you're right, if we do use negative values in our enum, we'd need to check for != 0.Rodmann
I
13

Define the Problem

Let’s define an enum that represents the types of users:

public enum UserType
{
    Customer = 1,
    Driver = 2,  
    Admin = 3,
}

We define the UserType enum that contains three values: Customer, Driver, and Admin.

But what if we need to represent a collection of values?

For example, at a delivery company, we know that both the Admin and the Driver are employees. So let’s add a new enumeration item Employee. Later on, we will show you how we can represent both the admin and the driver with it:

public enum UserType
{   
    Customer = 1,
    Driver = 2,  
    Admin = 3,
    Employee = 4
}

Define and Declare a Flags Attribute

A Flags is an attribute that allows us to represent an enum as a collection of values ​​rather than a single value. So, let’s see how we can implement the Flags attribute on enumeration:

[Flags]
public enum UserType
{   
    Customer = 1,
    Driver = 2,  
    Admin = 4,
}

We add the Flags attribute and number the values with powers of 2. Without both, this won’t work.

Now going back to our previous problem, we can represent Employee using the | operator:

var employee = UserType.Driver | UserType.Admin;

Also, we can define it as a constant inside the enum to use it directly:

[Flags]
public enum UserType
{                
    Customer = 1,             
    Driver = 2,               
    Admin = 4,                
    Employee = Driver | Admin
}

Behind the Scenes

To understand the Flags attribute better, we must go back to the binary representation of the number. For example, we can represent 1 as binary 0b_0001 and 2 as 0b_0010:

[Flags]
public enum UserType
{
    Customer = 0b_0001,
    Driver = 0b_0010,
    Admin = 0b_0100,
    Employee = Driver | Admin, //0b_0110
}

We can see that each value is represented in an active bit. And this is where the idea of ​​numbering the values ​​with the power of 2 came from. We can also note that Employee contains two active bits, that is, it is a composite of two values Driver and Admin.

Operations on Flags Attribute

We can use the bitwise operators to work with Flags.

Initialize a Value

For the initialization, we should use the value 0 named None, which means the collection is empty:

[Flags]
public enum UserType
{
    None = 0,
    Customer = 1,
    Driver = 2,
    Admin = 4,
    Employee = Driver | Admin
}

Now, we can define a variable:

var flags = UserType.None;

Add a Value

We can add value by using | operator:

flags |= UserType.Driver;

Now, the flags variable equals Driver.

Remove a Value

We can remove value by use &, ~ operators:

flags &= ~UserType.Driver;

Now, flagsvariable equals None.

We can check if the value exists by using & operator:

Console.WriteLine((flags & UserType.Driver) == UserType.Driver);

The result is False.

Also, we can do this by using the HasFlag method:

Console.WriteLine(flags.HasFlag(UserType.Driver));

Also, the result will be False.

As we can see, both ways, using the & operator and the HasFlag method, give the same result, but which one should we use? To find out, we will test the performance on several frameworks.

Measure the Performance

First, we will create a Console App, and in the .csproj file we will replace the TargetFramwork tag with the TargetFramworks tag:

<TargetFrameworks>net48;netcoreapp3.1;net6.0</TargetFrameworks>
We use the TargetFramworks tag to support multiple frameworks: .NET Framework 4.8, .Net Core 3.1, and .Net 6.0.

Secondly, let’s introduce the BenchmarkDotNet library to get the benchmark results:

[Benchmark]
public bool HasFlag()
{
    var result = false;
    for (int i = 0; i < 100000; i++)
    {
        result = UserType.Employee.HasFlag(UserType.Driver);
    }
    return result;
}
[Benchmark]
public bool BitOperator()
{
    var result = false;
    for (int i = 0; i < 100000; i++)
    {
        result = (UserType.Employee & UserType.Driver) == UserType.Driver;
    }
    return result;
}

We add [SimpleJob(RuntimeMoniker.Net48)], [SimpleJob(RuntimeMoniker.NetCoreApp31)], and [SimpleJob(RuntimeMoniker.Net60)] attributes to the HasFlagBenchmarker class to see the performance differences between different versions of .NET Framework / .NET Core:

Method Job Runtime Mean Error StdDev Median
HasFlag .NET 6.0 .NET 6.0 37.79 us 3.781 us 11.15 us 30.30 us
BitOperator .NET 6.0 .NET 6.0 38.17 us 3.853 us 11.36 us 30.38 us
HasFlag .NET Core 3.1 .NET Core 3.1 38.31 us 3.939 us 11.61 us 30.37 us
BitOperator .NET Core 3.1 .NET Core 3.1 38.07 us 3.819 us 11.26 us 30.33 us
HasFlag .NET Framework 4.8 .NET Framework 4.8 2,893.10 us 342.563 us 1,010.06 us 2,318.93 us
BitOperator .NET Framework 4.8 .NET Framework 4.8 38.04 us 3.920 us 11.56 us 30.17 us

So, in .NET Framework 4.8 a HasFlag method was much slower than the BitOperator. But, the performance improves in .Net Core 3.1 and .Net 6.0. So in newer versions, we can use both ways.

Intranuclear answered 4/1, 2023 at 15:27 Comment(1)
You don't need the [Flags] attribute to make this work. It will also work without. The only thing the [Flags] attribute does is change is the result of the ToString() of the enum. When you have an enum value Driver | Admin, without [Flags] it will print 6, and with [Flags] attribute it will print as Driver, Admin.Cold
E
1

Apologies if someone already noticed this scenario. A perfect example of flags we can see in reflection. Yes Binding Flags ENUM.

[System.Flags]
[System.Runtime.InteropServices.ComVisible(true)]
[System.Serializable]
public enum BindingFlags

Usage

// BindingFlags.InvokeMethod
// Call a static method.
Type t = typeof (TestClass);

Console.WriteLine();
Console.WriteLine("Invoking a static method.");
Console.WriteLine("-------------------------");
t.InvokeMember ("SayHello", BindingFlags.InvokeMethod | BindingFlags.Public | 
    BindingFlags.Static, null, null, new object [] {});
Egeria answered 27/1, 2020 at 22:51 Comment(0)
P
-12
  • Flags are used when an enumerable value represents a collection of enum members.

  • here we use bitwise operators, | and &

  • Example

                 [Flags]
                 public enum Sides { Left=0, Right=1, Top=2, Bottom=3 }
    
                 Sides leftRight = Sides.Left | Sides.Right;
                 Console.WriteLine (leftRight);//Left, Right
    
                 string stringValue = leftRight.ToString();
                 Console.WriteLine (stringValue);//Left, Right
    
                 Sides s = Sides.Left;
                 s |= Sides.Right;
                 Console.WriteLine (s);//Left, Right
    
                 s ^= Sides.Right; // Toggles Sides.Right
                 Console.WriteLine (s); //Left
    
Penance answered 11/11, 2020 at 7:59 Comment(1)
Your example is wrong (and does not produce the output it claims) because the proper use of Flags requires that your enum values have unique bits set. And there should be at least one set (unless you a pseudo value like None). Your enum should like ` public enum Sides { Left=1, Right=2, Top=4, Bottom=8 } `Fend

© 2022 - 2024 — McMap. All rights reserved.