The contents of a Byte
Array
of size 4 are the following: {1, 0, 0, 0}
. This translates to the integer number 1
in C# when using BitConverter.ToInt32(bytearray, 0);
However, when converting this byte array to an Integer
in Kotlin using the following code base I get the value 16777216
instead of the value of 1
.
var test0 = BigInteger(bytearray).toInt() = 16777216
var test1 = BigInteger(bytearray).toFloat() = 1.6777216
or
fun toInt32(bytes: ByteArray, index: Int): Int
{
if (bytes.size != 4)
throw Exception("The length of the byte array must be at least 4 bytes long.")
return 0xff
and bytes[index].toInt() shl 56
or (0xff and bytes[index + 1].toInt() shl 48)
or (0xff and bytes[index + 2].toInt() shl 40)
or (0xff and bytes[index + 3].toInt() shl 32)
}
I believe both methods of conversion are correct and the byte values are not signed.
As suggested by @Lother and itsme86.
fun littleEndianConversion(bytes: ByteArray): Int {
var result = 0
for (i in bytes.indices) {
result = result or (bytes[i].toInt() shl 8 * i)
}
return result
}
A tiny bit different solution which I think is a bit more efficient:
private fun byteToInt(bytes: ByteArray): Int {
var result = 0
var shift = 0
for (byte in bytes) {
result = result or (byte.toInt() shl shift)
shift += 8
}
return result
}
Here is,
fun toInt32(bytes:ByteArray):Int {
if (bytes.size != 4) {
throw Exception("wrong len")
}
bytes.reverse()
return ByteBuffer.wrap(bytes).int
}
Similar to eigenfeld's answer, but without needing to reverse the bytes first.
fun toInt32(bytes: ByteArray, index: Int): Int {
require(bytes.size == 4) { "length must be 4, got: ${bytes.size}" }
return ByteBuffer.wrap(bytes, index, 4).order(ByteOrder.LITTLE_ENDIAN).int
}
在 C# 中,您有以下类:
var test0 = BitConverter.ToInt32(bytearray, 0)
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