我正在阅读的源代码,EqualityComparer<T>.Default
发现它不是那么聪明。这是一个例子:
enum MyEnum : int { A, B }
EqualityComparer<MyEnum>.Default.Equals(MyEnum.A, MyEnum.B)
//is as fast as
EqualityComparer<int>.Default.Equals(0, 1)
enum AnotherEnum : long { A = 1L, B = 2L }
//is 8x slower than
EqualityComparer<long>.Default.Equals(1L, 2L)
原因从 EqualityComparer 中的私有方法的源码中就很明显了。
private static EqualityComparer<T> CreateComparer()
{
//non-important codes are ignored
if (c.IsEnum && (Enum.GetUnderlyingType(c) == typeof(int)))
{
return (EqualityComparer<T>) RuntimeTypeHandle.CreateInstanceForAnotherGenericParameter((RuntimeType) typeof(EnumEqualityComparer<int>), c);
}
return new ObjectEqualityComparer<T>();
}
我们可以看到EqualityComparer<int>.Default
,EqualityComparer<MyEnum>.Default
并EqualityComparer<long>.Default
得到一个明智的比较器,其Equals
方法如下:
public static bool Equals(int x, int y)
{
return x == y; //or return x.Equals(y); here
//I'm not sure, but neither causes boxing
}
public static bool Equals(MyEnum x, MyEnum y)
{
return x == y; //it's impossible to use x.Equals(y) here
//because that causes boxing
}
上面两个很聪明,但是EqualityComparer<AnotherEnum>.Default
很不幸,从方法中我们最后可以看到它得到一个ObjectEqualityComparer<T>()
,它的Equals
方法大概是这样的:
public static bool Equals(AnotherEnum x, AnotherEnum y)
{
return x.Equals(y); //too bad, the Equals method is from System.Object
//and it's not override, boxing here!
//that's why it's so slow
}
我认为这个条件Enum.GetUnderlyingType(c) == typeof(int)
是没有意义的,如果一个枚举的底层类型是 int 类型,该方法可以将int 的默认比较器转换为这个枚举。但是为什么不能基于 long 的枚举呢?我想这不是那么难吗?有什么特殊原因吗?构建一个比较器x == y
对于枚举来说并不难,对吧?为什么最后它会降低ObjectEqualityComparer<T>
枚举速度(即使它工作正常)?