byte array to unsigned int in java

Question

I\'m converting a byte array into int by doing this:

ByteArrayInputStream bais = new ByteArrayInputStream (data);
DataInputStream dis = new Data         


        

Answer 1

Java's primitive integer types (i.e. byte, short, int and long) are all signed. But you can work around this.

To do (say) 32 bit unsigned arithmetic, just do the arithmetic pretending that 'int' is unsigned. For example, 2**31 - 1 is the largest (signed) int value. If you add one to it you will get -2**31. But that bit pattern is the same as +2**31 if you think of the int as being unsigned. This also works for subtraction and multiplication. (I'm not sure about division and remainder, but the chances are that doesn't matter for you).

Comparing unsigned 32 bit values is a bit more tricky. For example, -1 is less that +1, but if you interpret -1 as an unsigned value you get +2**32 - 1 which should be greater than '+1'. You can compensate by translating the inequality (I'll leave it to the reader to figure it out) or by casting the int values to long, masking them with 0xffffffffL and comparing them as longs; e.g.

int u1 = ...
int u2 = ...
if ((((long) u1) & 0xffffffff) < (((long) u2) & 0xffffffff) {
    // u1 represents a smaller unsigned value than u2
}

Converting 32 bit unsigned integers to Strings is easiest done using longs; e.g.

String str = Long.toString(((long) u1) & 0xffffffffL);

Now I'll freely admit that using int to represent 32 bit unsigned values is tricky, and potentially error prone. A cleaner solution would be to use long throughout, or if your application needs 64 bit unsigned values to use BigInteger.

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UPDATE - it looks Java 8 will have support (in the form of library methods) for treating int and long as unsigned types - see "Unsigned Integer Arithmetic API now in JDK 8" by Joseph Darcy @ Oracle.

Answer 2

Each cell in the array is treated as unsigned int:

private int unsignedIntFromByteArray(byte[] bytes) {
    int res = 0;
    if (bytes == null)
        return res;

    for (int i = 0; i < bytes.length; i++) {
        res = (res *10) + ((bytes[i] & 0xff));
    }
    return res;
}

• The code will convert [12,34] to 1234

Answer 3

Java does not have unsigned int, so you'll have to manually invert the bits to get an unsigned value if you absolutely need to have one.

Got this from Google.

Answer 4

An int is always a signed, 32-bit number in Java. However, this only matters if you are doing math with it. If all you care about is the pattern of 0 and 1 bits, simply ignore the sign.

If you do need to do some math, convert it to a long by masking:

long l = j & 0xFFFFFFFFL;

Do all arithmetic with long operands, modulo 0xFFFFFFFFL. When you are done, cast the result back to an int and transmit it.