Best way to convert a signed integer to an unsigned long?

Question

For certain hash functions in Java it would be nice to see the value as an unsigned integer (e.g. for comparison to other implementations) but Java supports only signed types. We can convert a signed int to an "unsigned" long as such:

public static final int BITS_PER_BYTE = 8;
public static long getUnsignedInt(int x) {
  ByteBuffer buf = ByteBuffer.allocate(Long.SIZE / BITS_PER_BYTE);
  buf.putInt(Integer.SIZE / BITS_PER_BYTE, x);
  return buf.getLong(0);
}
getUnsignedInt(-1); // => 4294967295

However, this solution seems like overkill for what we're really doing. Is there a more efficient way to achieve the same thing?

score 109 · Accepted Answer · answered Mar 06 '12 at 06:00

109

Something like this?

int x = -1;
long y = x & 0x00000000ffffffffL;

Or am I missing something?

public static long getUnsignedInt(int x) {
    return x & 0x00000000ffffffffL;
}

answered Mar 06 '12 at 06:00

Mysticial

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@Paranaix I believe that would also work. It's been a while since I wrote this answer, so I think the reason why I padded with zeros was out of habit, since I do mostly C++ and the integer promotion rules there are more complicated with signed/unsigned types. – Mysticial Nov 29 '12 at 16:38

score 62 · Answer 2 · edited May 23 '17 at 12:10

62

The standard way in Java 8 is Integer.toUnsignedLong(someInt), which is equivalent to @Mysticial's answer.

edited May 23 '17 at 12:10

Community

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answered Jun 30 '15 at 12:37

superEb

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score 18 · Answer 3 · edited Jan 09 '18 at 19:12

18

Guava provides UnsignedInts.toLong(int)...as well as a variety of other utilities on unsigned integers.

edited Jan 09 '18 at 19:12

Iulian Popescu

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answered Mar 06 '12 at 06:37

Louis Wasserman

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score 9 · Answer 4 · answered Mar 06 '12 at 08:27

9

You can use a function like

public static long getUnsignedInt(int x) {
    return x & (-1L >>> 32);
}

however in most cases you don't need to do this. You can use workarounds instead. e.g.

public static boolean unsignedEquals(int a, int b) {
    return a == b;
}

For more examples of workarounds for using unsigned values. Unsigned utility class

answered Mar 06 '12 at 08:27

Peter Lawrey

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Your getUnsignedInt function doesn't work. I made a serious error in my program as I didn't test it. :(( – user2707175 Jul 08 '14 at 20:08
getUnsignedInt appears to work as-is. `>>>` is correct. `>>` causes sign extension, and -1L right-shifted with sign extension any amount always results in a value with all ones. – Macil Oct 18 '17 at 17:48
getUnsignedInt works for me too, as-is. – Simon Savai Jan 06 '22 at 11:33

score 2 · Answer 5 · answered Mar 06 '12 at 06:19

2

other solution.

public static long getUnsignedInt(int x) {
    if(x > 0) return x;
    long res = (long)(Math.pow(2, 32)) + x;
    return res;
}

answered Mar 06 '12 at 06:19

lmatt

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Math.pow is very expensive and calculated every time. By comparison `1l << 32` is fast and only calculated by the compiler. (And doesn't need to be cast to a long. – Peter Lawrey Mar 06 '12 at 08:23
Can you please give me some link or description on how these conversion work, m not able to understand it. – ManMohan Vyas Sep 28 '12 at 09:45
1

Because int is 32 bit and use two's complement to represent. You can check two's complement in wikipedia(http://en.wikipedia.org/wiki/Two's_complement) to get more details – lmatt Nov 14 '12 at 09:15

score -2 · Answer 6 · answered Mar 05 '15 at 11:27

-2

Just my 2 cents here, but I think it's a good practice to use:

public static long getUnsignedInt(int x) { return x & (~0L); // ~ has precedence over & so no real need for brackets }

instead of:

return x & 0xFFFFFFFFL;

In this situation there's not your concern how many 'F's the mask has. It shall always work!

answered Mar 05 '15 at 11:27

okoopat

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This is completely wrong. The int is promoted to a long, and then you're masking it with a bit string of 64 ones which does nothing. The following prints -1 instead of 4294967295: int i = -1; long j = i & (~0L); System.out.println(j); – PBJ Apr 23 '15 at 20:02
yes it's my bad, I confused 0xFFFFFFFFL with ~0L which obvious it's not the case – okoopat Jun 17 '15 at 12:10

score -4 · Answer 7 · answered Nov 19 '14 at 09:31

-4

long abs(int num){
    return num < 0 ? num * -1 : num;
}

answered Nov 19 '14 at 09:31

user2207488

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please read about two's complement (https://en.wikipedia.org/wiki/Two%27s_complement) which describes how signed numbers are usually represented. You cant just take absolute value for resolving that issue (http://tadej.me/twos-complement-and-absolute-values/). – Stepan Pogosyan Mar 18 '18 at 21:28

Best way to convert a signed integer to an unsigned long?

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