Combine 4 bytes into one in Java -

here code:

import java.io.fileoutputstream; import java.io.ioexception; import java.util.map;  public class shiftcodes {     private map<byte[], byte> shiftmap;      public byte genobyte(byte b1, byte b2, byte b3, byte b4) {         return (byte) (                   (b1 << 6)                 | (b2 << 4)                 | (b3 << 2)                 | b4);     }      public static void main(string[] args) throws ioexception {         shiftcodes shiftcodes = new shiftcodes();         byte b = shiftcodes.genobyte((byte) 0x01, (byte) 0x11, (byte) 0x00, (byte) 0x10);         fileoutputstream fileoutputstream = new fileoutputstream("/tmp/x.bin");         fileoutputstream.write(new byte[] {b});     } } 

it's assumed bits of each byte zero, except rightmost 2 bits, can 0 or 1. changed code little:

public class shiftcodes {     private map<byte[], byte> shiftmap;      public byte genobyte(byte b1, byte b2, byte b3, byte b4) {         return (byte) (                   ((b1 & 0x11) << 6)                 | ((b2 & 0x11) << 4)                 | ((b3 & 0x11) << 2)                 | b4);     }      public static void main(string[] args) throws ioexception {         shiftcodes shiftcodes = new shiftcodes();         byte b = shiftcodes.genobyte((byte) 0x01, (byte) 0x11, (byte) 0x00, (byte) 0x10);         fileoutputstream fileoutputstream = new fileoutputstream("/tmp/x.bin");         fileoutputstream.write(new byte[] {b});     } } 

but in both cases getting expected (01110010):

xxd -b x.bin 0000000: 01010000                                               p 

why?

you mistake hex literals binary literals:

0x11; // hex 11, dec 17, bits 00010001 0b11; // hex 3,  dec 3,  bits 00000011