Answer 1

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.

Answered by: Carlos670 | Posted: 24-02-2022

Answer 2

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

Answered by: Rafael191 | Posted: 24-02-2022

Answer 3

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);

Answered by: Max393 | Posted: 24-02-2022

Answer 4

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

Answered by: Julian472 | Posted: 24-02-2022

Answer 5

The answer of Jeremy is correct in case of positive integer value. If the conversion should be correct for negative values, it is little more complicated due to two's-complement format (https://en.wikipedia.org/wiki/Two%27s_complement). The trick is to remove the gap between the interesting bits (the less significant bits) and the 'sign' bit. One easy method is multiplication of the number.

    int myIntMultiplied = myInt * 256;

    byte b1, b2, b3;

    b3 = (byte) ((myIntMultiplied >> 8) & 0xFF);
    b2 = (byte) ((myIntMultiplied >> 16) & 0xFF);
    b1 = (byte) ((myIntMultiplied >> 24) & 0xFF);

This will end up with correct two's-complement format for negative values.

PS: You can check and compare binary representation this way:

Integer.toBinaryString(myInt);
Integer.toBinaryString(myIntMultiplied );

Answered by: Lydia131 | Posted: 24-02-2022

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