将有符号整数转换为无符号长整型的最佳方法？

Question

For certain hash functions in Java it would be nice to see the value as an unsigned integer (eg 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?

Answer 1

Something like this?

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

Or am I missing something?

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

Answer 2

Java 8中的标准方法是Integer.toUnsignedLong(someInt) ，这相当于@Mysticial的答案 。

Answer 3

Guava提供UnsignedInts.toLong(int) ...以及无符号整数上的各种其他实用程序。

Answer 4

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. eg

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

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

Answer 5

other solution.

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

Answer 6

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!

Answer 7

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