Convert hex string into long 17digits java

Question

I am trying to convert "0x6042607b1ba01d8dl" into a long.

I have tried:

long value = new BigInteger("0x6042607b1ba01d8dl", 16).longValue();
long value = new BigInteger("0x6042607b1ba01d8dl", 32).longValue();
long value = Long.decode("0x6042607b1ba01d8dl");
Long.parseLong("0x6042607b1ba01d8dl");

Note: The Hex number "0x6042607b1ba01d8dl" has 17 numbers

possible duplicate of [How to convert a hexadecimal string to long in java?](http://stackoverflow.com/questions/5153811/how-to-convert-a-hexadecimal-string-to-long-in-java) — turbo, May 21 '14 at 14:02
It would be quite helpful if you told us what happened as well, not just what you tried. — nos, May 21 '14 at 14:19
"The Hex number ... has 17 numbers." Unless I'm mistaken, `6042 607b 1ba0 1d8d` has **16** digits so should fit in a 64b Java `long`. — Mike Samuel, May 21 '14 at 14:23

score 3 · Answer 1 · edited May 21 '14 at 14:26

3

From the javadoc for the BigInteger(String,int) constructor:

The String representation consists of an optional minus or plus sign followed by a sequence of one or more digits in the specified radix.

So you just need to remove the 0x from your string:

long value = new BigInteger("6042607b1ba01d8d", 16).longValue();

edited May 21 '14 at 14:26

Mike Samuel

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answered May 21 '14 at 14:03

Giovanni Botta

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score 2 · Answer 2 · answered May 21 '14 at 14:03

2

The BigInteger constructor does not understand your 0x prefix.

Use e.g.

long value = new BigInteger("6042607b1ba01d8d", 16).longValue();

Or:

String number = "0x6042607b1ba01d8d";
long value = new BigInteger(number.subString(2), 16).longValue();

You can also use Long.decode(), which does accept a 0x prefix for decoding hex.

answered May 21 '14 at 14:03

nos

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OP did use `Long.decode("0x6042607b1ba01d8d");` I'm not sure about the outcome tho – Goodwine May 21 '14 at 14:12

score 1 · Answer 3 · edited May 22 '14 at 14:22

1

You can try this:

long value = Long.parseLong("6042607b1ba01d8d", 16);

Long.parseLong can sometimes fail for unsigned longs, so, the BigInteger approaches are better.

edited May 22 '14 at 14:22

O. Jones

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answered May 21 '14 at 14:06

Pablo Arce

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Welcome to StackOverflow. Answers with some explanation are preferable. – O. Jones May 21 '14 at 15:26
Thanks. I have seen that Long.parseLong can fail for unsigned longs, so, the BigInteger approches are better. – Pablo Arce May 22 '14 at 14:20

score 0 · Answer 4 · answered May 21 '14 at 14:08

0

As said by the above answers in code form:

String bigHexNumber = "0x6042607b1ba01d8d";
if(bigHexNumber.subString(0, 1).equals("0x") {
     bigHexNumber = bigHexNumber.subString(2);
}
long hexInLongForm = new BigInteger(bigHexNumber, 16).longValue();

answered May 21 '14 at 14:08

DirkyJerky

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Convert hex string into long 17digits java

4 Answers4