如何在Java中将十六进制字符串转换为long？

Question

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!

Answer 1

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

Answer 2

使用parseLong：

Long.parseLong(s, 16)

Answer 3

new BigInteger(string, 16).longValue()

For any value of someLong:

new BigInteger(Long.toHexString(someLong), 16).longValue() == someLong

In other words, this will return the long you sent into Long.toHexString() for any long value, including negative numbers. It will also accept strings that are bigger than a long and silently return the lower 64 bits of the string as a long . You can just check the string length <= 16 (after trimming whitespace) if you need to be sure the input fits in a long .

Answer 4

Long.parseLong(s, 16) will only work up to "7fffffffffffffff" . Use BigInteger instead:

public static boolean isHex(String hex) {
    try {
        new BigInteger(hex, 16);
        return true;
    } catch (NumberFormatException e) {
        return false;
    }
}