Question

按照我的测试来源进行操作。从枚举对象获取值的好方法是什么？必须支持多久。我正在尝试没有try / catch块。

enum ELong: long { a = 0x100000000 };
enum ENormal { a = 25 }

        var l = (object) ELong.a;
        var n = (object)ENormal.a;
        //will cast into the correct size
        int ii = (int)n; //ok
        long ll = (long)l; //ok
        //wont cast if its too big
        ll = (long)n; //cast exception
        //or too small
        n = (int)l; //cast exception//compile error. Cannot cast
        //lets try preventing the exception with is
        if (n is int)
            ii = (int)n;//doesnt get here.
        if (n is long)
            ll = (long)n;//doesnt get here.
        if (l is int)
            ii = (int)l;//doesnt get here
        if (l is long)
            ll = (long)l;//doesnt get here
        //WHY!!!!
        //Maybe as will do the trick?
        if (n as int? != null)
            ii = (int)n;//doesnt get here.
        if (n as long? != null)
            ll = (long)n;//doesnt get here.
        if (l as int? != null)
            ii = (int)l;//doesnt get here
        if (l as long? != null)
            ll = (long)l;//doesnt get here
        //geez. What is more stange is (int) will work while (int?) will not
        int? ni = (int?)n;//cast exception
        int iii = (int)n; //works
        ll = (long)n;

Answer 1

long test1 = Convert.ToInt64(l); // 4294967296
long test2 = Convert.ToInt64(n); // 25

Answer 2

<强>解释

if (n is int)
        ii = (int)n;//doesnt get here.
    if (n is long)
        ll = (long)n;//doesnt get here.
    if (l is int)
        ii = (int)l;//doesnt get here
    if (l is long)
        ll = (long)l;//doesnt get here

n 和 l 既不是 int / long 也不是 long？/ int？，它们属于类型你的枚举，所以这是预期的行为。

<强>解决方案

可能你应该使用转换类来实现你想要的目标。

Answer 3

来自MSDN的示例：

static object GetAsUnderlyingType(Enum enval)
{
    Type entype = enval.GetType();

    Type undertype = Enum.GetUnderlyingType(entype);

    return Convert.ChangeType( enval, undertype );
}

转换为基础类型（int或long）
将转换后的值转换为您的变量（我想简单地将结果分配给长变量）。

将enum从object转换为long而不使用try块？

3 个答案: