16进制字符串怎么转换为对应的ASCII码

16进制字符串如何转换为对应的ASCII码？
16进制的字符串如何转换成对应的ASCII码？

CString hexstr = "30";

string ascii = xxxx(hexstr);//有没有.NET框架的系统函数可以转换的呢？谢谢！不会自己建个内存表来查找吧？


如何得到对应的ASCII码？


[解决办法]

C# code

class Program{  static void Main()  {    string hexstr = "30";    string ascii = ((char)int.Parse(hexstr, System.Globalization.NumberStyles.HexNumber)).ToString();      System.Console.WriteLine(ascii);  }}
[解决办法]
C# code
string str = "30";int sss=Convert.ToInt32(str,16);
[解决办法]
给你一段完整的代码:
       const string HexChs = "0123456789ABCDEF";


       private static string str2hex(string s)
       {
           StringBuilder sb = new StringBuilder();
           for (int i=0;i<s.Length; i++) {
               if (i>0) sb.Append(' ');
               sb.Append( ((UInt16)s[i]).ToString("X2"));
           }

           return sb.ToString();
       }

       private static string hex2str(string s)
       {
           string str = s.ToUpper();
           StringBuilder sb = new StringBuilder();
           for (int i=0; i<str.Length; i+=3){
               int n1 = HexChs.IndexOf( str[i]);
               if (n1==-1) throw new Exception("格式错");
               if (i+1==str.Length) throw new Exception("格式错");
               int n2 = HexChs.IndexOf( str[i+1]);
               if (n2==-1) throw new Exception("格式错");
               if (i+2< str.Length && str[i+2]!= ' ') throw new Exception("格式错");
               char c = (char)((n1 << 4) + n2);
               sb.Append(c);
           }
           return sb.ToString();
       }

[解决办法]
探讨

反过来怎么算

知道 ASCII 0 怎么转换到 16进制30去？

[解决办法]
Java code
public static String toStringHex(String s)     {     byte[] baKeyword = new byte[s.length()/2];     for(int i = 0; i < baKeyword.length; i++)     {     try     {     baKeyword[i] = (byte)(0xff & Integer.parseInt(s.substring(i*2, i*2+2),16));     }     catch(Exception e)     {     e.printStackTrace();     }     }     try     {     s = new String(baKeyword, "utf-8");// UTF-16le:Not    }     catch (Exception e1)     {     e1.printStackTrace();     }     return s;     }