使用 ByteBuffer 将 n 字节数组转换为 integer

Question

I need to convert arbitrary sized byte arrays to a short / int / long .
This means I could receive 3 bytes to be converted to int :

final byte[] bytes = { 0b0000, 0b0000, 0b1111 }; // 15 - big endian
final int value = doConversion(bytes);

Thus, I'm trying to come up with a generic function.

ByteBuffer conversions works great when you have arrays of a size that exactly represent a short , int , long value. But what if I have an int represented as a single byte?

final byte[] bytes = { 0b1111 }; // 15

It seems converting such a byte array to an int using a ByteBuffer requires resizing the array and padding it.
Things get even more complicated with a negative value, as the padding needs to be done with the most significant bit .

final byte[] bytes = { (byte) 0b11110001 }; // -15 stored as two's complement

Is there an easier way to accomplish this task, or should I just use custom code?
An example could be, in Kotlin, using extension functions:

fun ByteArray.toShort(byteOrder: ByteOrder = LITTLE, signed: Boolean = true): Short =
    toInteger(Short.SIZE_BYTES, byteOrder, signed).toShort()

fun ByteArray.toInt(byteOrder: ByteOrder = LITTLE, signed: Boolean = true): Int =
    toInteger(Int.SIZE_BYTES, byteOrder, signed).toInt()

fun ByteArray.toLong(byteOrder: ByteOrder = LITTLE, signed: Boolean = true): Long =
    toInteger(Long.SIZE_BYTES, byteOrder, signed)

private fun ByteArray.toInteger(
     typeBytes: Int /* 2, 4, 8 */,
     byteOrder: ByteOrder /* little, big */, 
     signed: Boolean,
): Long {
    // Checks omitted...

    // If the byte array is bigger than the type bytes, it needs to be truncated
    val bytes =
        if (size > typeBytes) {
            if (byteOrder == LITTLE) {
                copyOf(typeBytes)
            } else {
                copyOfRange(size - typeBytes, size)
            }
        } else {
            copyOf()
        }

    val negative =
        signed && (this[if (byteOrder == LITTLE) bytes.size - 1 else 0]).toInt() and 0b1000000 != 0

    if (!negative) {
        return bytes.absoluteToLong(byteOrder)
    }

    // The number is stored as two's complement.
    // Thus we invert each byte and then sum 1 to obtain the absolute value
    for (i in bytes.indices) {
        bytes[i] = bytes[i].inv()
    }

    return -(bytes.absoluteToLong(byteOrder) + 1)
}

private fun ByteArray.absoluteToLong(byteOrder: ByteOrder): Long {
    var result = 0L
    var shift = 8 * (size - 1)
    val range =
        if (byteOrder == LITTLE) {
            size - 1 downTo 0
        } else {
            0 until size
        }

    for (i in range) {
        result = (this[i].toInt() and 0b11111111).toLong() shl shift or result
        shift -= 8
    }

    return result
}

Answer 1

The BigInteger class has a constructor that's convenient for this use case. Example:

byte[] bytes = { 0b1111 }; 
int value = new BigInteger(bytes).intValue(); 

It does "sign-extension" for you: if the first byte value in the array is negative, the end result is negative.

It has also methods to retrieve the value as the other number types (short, long, ...).