跟随我的测试源。从枚举对象获取值的好方法是什么?必须长期支持。我正在尝试不使用 try/catch block 。
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;
最佳答案
long test1 = Convert.ToInt64(l); // 4294967296
long test2 = Convert.ToInt64(n); // 25
关于c# - 在没有 try block 的情况下将枚举从对象转换为 long?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/2064028/