For my project, I want to convert an integer to an array of bytes and an array of bytes back to an integer

Question

For my project, I need to convert an integer to an array of bytes and an array of bytes back to an integer. I made the code below, but it does not work

int temp=200; byte[] tempbyte= new byte[4]; String s= String.valueOf(temp); tempbyte= s.getBytes(); // integer to byte array conversion String s2 = String.valueOf(ByteBuffer.wrap(tempbyte).getInt()); int temp2 = Integer.parseInt(s2); // byte array to integer conversion System.out.println("the value of temp2 = "+temp2); 

Answer 1

Abstracts are functions that I did many years ago that can help you:

 public static float byteArrayToFloat(byte[] p_bArray) { int nValue; nValue = (0xFF & p_bArray[0]); nValue = nValue | ((0xFF & p_bArray[1]) << 8); nValue = nValue | ((0xFF & p_bArray[2]) << 16); nValue = nValue | ((0xFF & p_bArray[3]) << 24); float fValue = Float.intBitsToFloat(nValue); return fValue; } public static byte[] floatToByteArray(float p_fValue) { byte [] bArray = new byte[4]; int nValue; nValue = Float.floatToIntBits(p_fValue); bArray[0] = (byte)nValue; bArray[1] = (byte)(nValue >> 8); bArray[2] = (byte)(nValue >> 16); bArray[3] = (byte)(nValue >> 24); return bArray; } public static double byteArrayToDouble(byte[] p_bArray) { long lValue; lValue = (long)p_bArray[0] & 0xFF; lValue = lValue | (((long)p_bArray[1] & 0xFF) << 8); lValue = lValue | (((long)p_bArray[2] & 0xFF) << 16); lValue = lValue | (((long)p_bArray[3] & 0xFF) << 24); lValue = lValue | (((long)p_bArray[4] & 0xFF) << 32); lValue = lValue | (((long)p_bArray[5] & 0xFF) << 40); lValue = lValue | (((long)p_bArray[6] & 0xFF) << 48); lValue = lValue | (((long)p_bArray[7] & 0xFF) << 56); double dValue = Double.longBitsToDouble(lValue); return dValue; } public static byte[] doubleToByteArray(double p_dValue) { byte [] bArray = new byte[8]; long lValue; lValue = Double.doubleToLongBits(p_dValue); bArray[0] = (byte)lValue; bArray[1] = (byte)(lValue >> 8); bArray[2] = (byte)(lValue >> 16); bArray[3] = (byte)(lValue >> 24); bArray[4] = (byte)(lValue >> 32); bArray[5] = (byte)(lValue >> 40); bArray[6] = (byte)(lValue >> 48); bArray[7] = (byte)(lValue >> 56); return bArray; } public static int byteArrayToInt(byte[] p_bArray) { int nValue; nValue = ((int)p_bArray[0]) & 0x000000FF; nValue = nValue | (((int)p_bArray[1] & 0x000000FF) << 8); nValue = nValue | (((int)p_bArray[2] & 0x000000FF) << 16); nValue = nValue | (((int)p_bArray[3] & 0x000000FF) << 24); return nValue; } public static byte[] intToByteArray(int p_nValue) { byte [] bArray = new byte[4]; bArray[0] = (byte)p_nValue; bArray[1] = (byte)(p_nValue >> 8); bArray[2] = (byte)(p_nValue >> 16); bArray[3] = (byte)(p_nValue >> 24); return bArray; } 

Answer 2

Try something like this: -

 byte[] arr = { 0x00, 0x01 }; ByteBuffer wrapped = ByteBuffer.wrap(arr); short num = wrapped.getShort(); ByteBuffer dbuf = ByteBuffer.allocate(2); dbuf.putShort(num); byte[] bytes = dbuf.array(); 

Use java.nio namespace classes

Answer 3

My version:

  int i = 1; byte[] bytes = BigInteger.valueOf(i).toByteArray(); i = new BigInteger(bytes).intValue(); 

Answer 4

The Integer class has a method for converting a number to a byte representation.

  String i = Integer.toBinaryString(4); 

and you can also do the opposite

  String binaryString = "100"; int base = 2; int decimal = Integer.parseInt(binaryString, base);