How to convert a hexadecimal string to long in java?

问题

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!

回答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

回答2:

Use parseLong:

Long.parseLong(s, 16)

回答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.

回答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;
    }
}

来源：https://stackoverflow.com/questions/5153811/how-to-convert-a-hexadecimal-string-to-long-in-java