假设我有这样一个enum:

[Flags]
enum Letters
{
     A = 1,
     B = 2,
     C = 4,
     AB = A | B,
     All = A | B | C,
}

为了检查AB是否被设置,我可以这样做:

if((letter & Letters.AB) == Letters.AB)

有没有一种比下面更简单的方法来检查是否设置了一个组合标志常量的任何标志?

if((letter & Letters.A) == Letters.A || (letter & Letters.B) == Letters.B)

例如,可以将&与其他东西交换吗?


当前回答

在. net 4中,您可以使用Enum。HasFlag方法:

using System;

[Flags] public enum Pet {
   None = 0,
   Dog = 1,
   Cat = 2,
   Bird = 4,
   Rabbit = 8,
   Other = 16
}

public class Example
{
   public static void Main()
   {
      // Define three families: one without pets, one with dog + cat and one with a dog only
      Pet[] petsInFamilies = { Pet.None, Pet.Dog | Pet.Cat, Pet.Dog };
      int familiesWithoutPets = 0;
      int familiesWithDog = 0;

      foreach (Pet petsInFamily in petsInFamilies)
      {
         // Count families that have no pets. 
         if (petsInFamily.Equals(Pet.None))
            familiesWithoutPets++;
         // Of families with pets, count families that have a dog. 
         else if (petsInFamily.HasFlag(Pet.Dog))
            familiesWithDog++;
      }
      Console.WriteLine("{0} of {1} families in the sample have no pets.", 
                        familiesWithoutPets, petsInFamilies.Length);   
      Console.WriteLine("{0} of {1} families in the sample have a dog.", 
                        familiesWithDog, petsInFamilies.Length);   
   }
}

使用实例显示如下信息:

//       1 of 3 families in the sample have no pets. 
//       2 of 3 families in the sample have a dog.

其他回答

在. net 4中,您可以使用Enum。HasFlag方法:

using System;

[Flags] public enum Pet {
   None = 0,
   Dog = 1,
   Cat = 2,
   Bird = 4,
   Rabbit = 8,
   Other = 16
}

public class Example
{
   public static void Main()
   {
      // Define three families: one without pets, one with dog + cat and one with a dog only
      Pet[] petsInFamilies = { Pet.None, Pet.Dog | Pet.Cat, Pet.Dog };
      int familiesWithoutPets = 0;
      int familiesWithDog = 0;

      foreach (Pet petsInFamily in petsInFamilies)
      {
         // Count families that have no pets. 
         if (petsInFamily.Equals(Pet.None))
            familiesWithoutPets++;
         // Of families with pets, count families that have a dog. 
         else if (petsInFamily.HasFlag(Pet.Dog))
            familiesWithDog++;
      }
      Console.WriteLine("{0} of {1} families in the sample have no pets.", 
                        familiesWithoutPets, petsInFamilies.Length);   
      Console.WriteLine("{0} of {1} families in the sample have a dog.", 
                        familiesWithDog, petsInFamilies.Length);   
   }
}

使用实例显示如下信息:

//       1 of 3 families in the sample have no pets. 
//       2 of 3 families in the sample have a dog.

抱歉,但我将在VB中显示它:)

   <Flags()> Public Enum Cnt As Integer
        None = 0
        One = 1
        Two = 2
        Three = 4
        Four = 8    
    End Enum

    Sub Test()
    Dim CntValue As New Cnt
    CntValue += Cnt.One
    CntValue += Cnt.Three
    Console.WriteLine(CntValue)
    End Sub

CntValue = 5 所以enum包含1 + 4

if((int)letter != 0) { }

如果可以使用。net 4或更高版本,请使用HasFlag()方法

例子

letter.HasFlag(Letters.A | Letters.B) // both A and B must be set

一样

letter.HasFlag(Letters.AB)

你可以检查这个值是否不为零。

if ((Int32)(letter & Letters.AB) != 0) { }

但是我认为引入一个值为0的新枚举值并与该枚举值进行比较是更好的解决方案(如果可能的话,因为您必须能够修改枚举)。

[Flags]
enum Letters
{
    None = 0,
    A    = 1,
    B    = 2,
    C    = 4,
    AB   =  A | B,
    All  = AB | C
}

if (letter != Letters.None) { }

更新

看错了问题——修正了第一个建议,忽略第二个建议。