C# switch 语句的默认标签将如何处理可为空的枚举?
默认标签会捕获空值和任何未处理的情况吗?
如果它为空,它将命中默认标签。
public enum YesNo
{
Yes,
No,
}
public class Program
{
public static void Main(string[] args)
{
YesNo? value = null;
switch (value)
{
case YesNo.Yes:
Console.WriteLine("Yes");
break;
case YesNo.No:
Console.WriteLine("No");
break;
default:
Console.WriteLine("default");
break;
}
}
}
该程序将打印default
.
除非 null 被处理。
public class Program
{
public static void Main(string[] args)
{
YesNo? value = null;
switch (value)
{
case YesNo.Yes:
Console.WriteLine("Yes");
break;
case YesNo.No:
Console.WriteLine("No");
break;
case null:
Console.WriteLine("NULL");
break;
default:
Console.WriteLine("default");
break;
}
}
}
打印NULL
。
如果您有稍后添加的未处理枚举值:
public enum YesNo
{
Yes,
No,
FileNotFound,
}
public class Program
{
public static void Main(string[] args)
{
YesNo? value = YesNo.FileNotFound;
switch (value)
{
case YesNo.Yes:
Console.WriteLine("Yes");
break;
case YesNo.No:
Console.WriteLine("No");
break;
default:
Console.WriteLine("default");
break;
}
}
}
它仍然打印default
。
您可以使用 null-coalescing 运算符??
将开关值路由null
到特定案例标签,而不是default
:
public static IEnumerable<String> AsStrings(this IEnumerable<Char[]> src)
{
Char[] rgch;
var e = src.GetEnumerator();
while (e.MoveNext())
{
switch ((rgch = e.Current)?.Length ?? -1)
{
case -1: // <-- value when e.Current is 'null'
yield return null;
break;
case 0:
yield return String.Empty;
break;
case 1:
yield return String.Intern(new String(rgch[0], 1));
break;
default: // 2...n
yield return new String(rgch);
break;
}
}
}
值得一提的是,C# 8.0 为 switch 表达式引入了新的属性模式。现在您可以使用下划线实现默认逻辑切换:
public double Calculate(int left, int right, Operator op) =>
op switch
{
Operator.PLUS => left + right,
Operator.MINUS => left - right,
Operator.MULTIPLY => left * right,
Operator.DIVIDE => left / right,
_ => 0 // default
}
参考。https://docs.microsoft.com/en-us/dotnet/csharp/whats-new/csharp-8