Best way to convert a signed integer to an unsigned long?
JavaUnsignedJava Problem Overview
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?
Java Solutions
Solution 1 - Java
Something like this?
int x = -1;
long y = x & 0x00000000ffffffffL;
Or am I missing something?
public static long getUnsignedInt(int x) {
return x & 0x00000000ffffffffL;
}
Solution 2 - Java
The standard way in Java 8 is Integer.toUnsignedLong(someInt)
, which is equivalent to @Mysticial's answer.
Solution 3 - Java
Guava provides UnsignedInts.toLong(int)
...as well as a variety of other utilities on unsigned integers.
Solution 4 - Java
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
Solution 5 - Java
other solution.
public static long getUnsignedInt(int x) {
if(x > 0) return x;
long res = (long)(Math.pow(2, 32)) + x;
return res;
}
Solution 6 - Java
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!
Solution 7 - Java
long abs(int num){
return num < 0 ? num * -1 : num;
}