Long.decode（str） 接受多种格式：

接受十进制，十六进制，以及由以下
语法给出的八进制
数字：

DecodableString：

• 签名 _opt DecimalNumeral


• Sign _opt 0x HexDigits


• Sign _opt 0X HexDigits
  
• Sign _opt ＃HexDigits


• Sign _opt 0 OctalDigits
>

签名：

• -

但是在你的情况下，这并不会起作用，你的字符串已经超出了长期持有的范围。您需要 BigInteger ：

  String s =4d0d08ada45f9dde1e99cad9; 
 BigInteger bi = new BigInteger（s，16）; 
 System.out.println（bi）; 
 

输出：

23846102773961507302322850521

对于比较，这里是 Long .MAX_VALUE ：

9223372036854775807

I want to convert a hex string to long in java.

I have tried with general conversion.

String s = "4d0d08ada45f9dde1e99cad9";
long l = Long.valueOf(s).longValue();
System.out.println(l);
String ls = Long.toString(l);

But I am getting this error message:

java.lang.NumberFormatException: For input string: "4d0d08ada45f9dde1e99cad9"

Is there any way to convert String to long in java? Or am i trying which is not really possible!!

Thanks!

Long.decode(str) accepts a variety of formats:

Accepts decimal, hexadecimal, and octal numbers given by the following grammar:
DecodableString:

• Sign_opt DecimalNumeral
• Sign_opt 0x HexDigits
• Sign_opt 0X HexDigits
• Sign_opt # HexDigits
• Sign_opt 0 OctalDigits

Sign:

• -

But in your case that won't help, your String is beyond the scope of what long can hold. You need a BigInteger:

String s = "4d0d08ada45f9dde1e99cad9";
BigInteger bi = new BigInteger(s, 16);
System.out.println(bi);

Output:

23846102773961507302322850521

For Comparison, here's Long.MAX_VALUE:

9223372036854775807