Question

I am trying to convert an int into three bytes representing that int (big endian).

I'm sure it has something to do with bit-wise and and bit shifting. But I have no idea how to go about doing it.

For example:

int myInt;

// some code

byte b1, b2 , b3; // b1 is most significant, then b2 then b3.

*Note, I am aware that an int is 4 bytes and the three bytes have a chance of over/underflowing.

Was it helpful?

Solution

To get the least significant byte:

b3 = myInt & 0xFF;

The 2nd least significant byte:

b2 = (myInt >> 8) & 0xFF;

And the 3rd least significant byte:

b1 = (myInt >> 16) & 0xFF;

Explanation:

Bitwise ANDing a value with 0xFF (11111111 in binary) will return the least significant 8 bits (bits 0 to 7) in that number. Shifting the number to the right 8 times puts bits 8 to 15 into bit positions 0 to 7 so ANDing with 0xFF will return the second byte. Similarly, shifting the number to the right 16 times puts bits 16 to 23 into bit positions 0 to 7 so ANDing with 0xFF returns the 3rd byte.

OTHER TIPS

byte b1 = (myint >> 16) & 0xff;
byte b2 = (myint >> 8) & 0xff;
byte b3 = myint & 0xff;

I am unsure how this holfds in java though, i aam not a java dev

An int doesn't fit into 3 bytes. However, assuming that you know these particular ones do:

   byte b1 = (myInt & 0xff);
   myInt >>= 8;
   byte b2 = (myInt & 0xff);
   myInt >>= 8;
   byte b3 = (myInt & 0xff);

In Java

int myInt = 1;
byte b1,b2,b3;
b3 = (byte)(myInt & 0xFF);
b2 = (byte)((myInt >> 8) & 0xFF);
b1 = (byte)((myInt >> 16) & 0xFF);
System.out.println(b1+" "+b2+" "+b3);

outputs 0 0 1

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