如何在Java中为幻数设置4字节位模式？

Question

I am working on a network programming on linux machine using C++ and I was wondering how I can set 4 bytes bit pattern for magic number in Java (client). Also how do I set the same bit pattern in c++ co I can compare the one from client with the one on the server side.

Thank in advance..

Edit

So now I have this

    byte[] signature = new byte[4];

    for(int i=0; i<4; i++){
        signature[i] = (byte) 0xA9;
    }

And if I looked at the inside of the array after for loop from the debugger then I have

{-89, -89, -89, -89}

And I did something like this in C++

    uint8_t m_magicNumberBuffer[4];
    magicKeyRead = read(m_fd, m_magicNumberBuffer, SIZE_OF_HEADER);


    if(m_magicNumberBuffer[0] == 0xA9 && m_magicNumberBuffer[1] == 0xA9 && m_magicNumberBuffer[2] == 0xA9 && m_magicNumberBuffer[3] == 0xA9){
        printf("SocketClient::recvMagicKey, Magic key has been found \n");
        break;
    }

I somehow works but not sure that I have declared m_magicNumberBuffer and unsigned integer but those were in negative 89 in java. Is this ok to do this in this way?

Thanks in advance.

Answer 1

Java has bitwise operators, for example bitwise OR | , bitwise AND & and bit shift operators >>> , >> and << , very similar to what C++ has. You can use those to manipulate bits exactly as you want.

Since you don't explain in more detail what you want to do, I cannot give you a more detailed answer.

Answer 2

In Java, you would represent it as

byte[] signature=new byte[4];

in C++, it would be

uint8_t signature[4];

You can then access each of the bytes individually as elements of the array.

Both languages support hex codes, so for example, you could do

signature[0]=0xA9;

in either java or C++ and it will set the first bit to A9 in hexadecimal (which is 10101001 in binary)

Answer 3

When you write to a DataOutputStream you write 8-bit bytes (sign is not important)

String filename = "text.dat";
DataOutputStream dos = new DataOutputStream(new FileOutputStream(filename));
for (int i = 0; i < 4; i++)
    dos.write(0xA9);
dos.close();

DataInputStream dis = new DataInputStream(new FileInputStream(filename));
for (int i = 0; i < 4; i++)
    System.out.println(Integer.toHexString(dis.readUnsignedByte()));
dis.close();

prints

a9
a9
a9
a9

Java assumes a byte is signed by default, however its is just 8-bits of data and used correctly can be unsigned or mean whatever you want it to.

Answer 4

What you're really dealing with is four bytes of raw memory; all you're concerned with is the bit pattern, not the numeric values. In eight bits (the size of a byte in Java), -89 and 0xA9 both have the same bit pattern: 10101001 . Because byte is signed in Java, dumping the value will show a negative value, which is rather counter intuitive, but Java doesn't have an eight bit unsigned type.

(Technically, 0xA9 isn't representable in a byte , and trying to put it in a signed char in C++ is illegal. But Java doesn't care about such niceties.)