How to convert a hexadecimal string to long in java?

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!

Solution 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

Solution 2

Use parseLong:

Long.parseLong(s, 16)

Solution 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.

Solution 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;
    }
}

View more solutions

Comments

[Sean Patrick Floyd]

True in general, but fails in this case (the number is much too large for a Long)

[Dead Programmer]

still it throws NumberFormatException,it's too long.

[Erik]

@Suresh: Yes, your sample string cannot be converted to a long, I was just answering "How to convert a hexadecimal string to long". You may want to look at BigInteger as Sean Patrick Floyd mentioned.

[LCoelho]

The goal is to convert to a Long, right? Then this is the clean and easy way to do it!

[J E Carter II]

Not sure about System.out.println method signature, but using bi in a String assignment I needed to use bi.toString()

[Sean Patrick Floyd]

@JECarterII println on an Object calls that object's toString() method

[user207421]

@SeanPatrickFloyd That's the OP's problem, not this answer's.

[JonSlowCN]

@SeanPatrickFloyd If the string cannot be converted to a Long, it should fail and throw an exception. It's the developer's job to handle that exception.

[ashdnik]

Crisp Clear Thanks!

[PLG]

By printing the stack trace, it seems that Long.decode() just calls Long.parseLong().

[Sean Patrick Floyd]

@PLG it's a facade around Long.valueOf, which in turn uses Long.parseLong. The specific purpose of Long.decode is to detect whether the input string is decimal, octal, or hexadecimal, and then call Long.valueOf with the detected base.

[Coronel Kittycannon]

This is what no-one is saying. Technically, you should be able to parse a unsigned long with a string with lenght 16 and only F's (FFFFFFFFFFFFFFFF), but Long.parseLong() only works up to 7fffffffffffffff, like you said.