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!

Try to add a "0x" prefix to the string before calling valueOf — sinelaw, Mar 01 '11 at 11:27
Umm ... did you look in the javadocs? Save yourself time and always look in the javadocs first. — Stephen C, Mar 01 '11 at 11:27
Also, valueOf should accept a second parameter that specifies the base. Try passing 16 (and the string without the prefix) — sinelaw, Mar 01 '11 at 11:28
@sinelaw - I think you are confusing `Long.valueOf(String)` with `Long.decode(String)`. The Javadoc for `valueOf(String)` says *"Parses the string argument as a signed decimal long."* — Stephen C, Mar 01 '11 at 11:32
Stephen, right, it's decode that accepts the prefix. But valueOf accepts the base parameter (see link above) — sinelaw, Mar 01 '11 at 11:39

Sean Patrick Floyd · Accepted Answer · 2011-03-01T12:08:31.580

100

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

edited Mar 01 '11 at 12:08

answered Mar 01 '11 at 11:36

Sean Patrick Floyd

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Not sure about System.out.println method signature, but using bi in a String assignment I needed to use bi.toString() – J E Carter II Apr 01 '15 at 16:51
2

@JECarterII println on an Object calls that object's toString() method – Sean Patrick Floyd Oct 15 '15 at 20:30
Crisp Clear Thanks! – ashdnik Jun 27 '18 at 12:38
By printing the stack trace, it seems that `Long.decode()` just calls `Long.parseLong()`. – PLG Nov 06 '19 at 10:48
@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. – Sean Patrick Floyd Nov 11 '19 at 17:48

score 47 · Answer 2 · answered Mar 01 '11 at 11:24

47

Use parseLong:

Long.parseLong(s, 16)

answered Mar 01 '11 at 11:24

Erik

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4

True in general, but fails in this case (the number is much too large for a Long) – Sean Patrick Floyd Mar 01 '11 at 11:43
still it throws NumberFormatException,it's too long. – Dead Programmer Mar 01 '11 at 11:55
@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. – Erik Mar 01 '11 at 12:04
1

The goal is to convert to a Long, right? Then this is the clean and easy way to do it! – LCoelho Nov 28 '14 at 13:34
@SeanPatrickFloyd That's the OP's problem, not this answer's. – user207421 Jul 18 '16 at 03:50
@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. – JonSlowCN Jan 05 '18 at 02:31

score 15 · Answer 3 · edited Apr 14 '12 at 23:36

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.

score 10 · Answer 4 · answered Mar 06 '15 at 11:49

10

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;
    }
}

answered Mar 06 '15 at 11:49

quezacoatl

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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. – Coronel Kittycannon Dec 20 '19 at 19:21

How to convert a hexadecimal string to long in java?

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