LSB/MSB handling in Java

Question

If I have to handle values to be stored in bytes like 0x118, how do I split the LSB and MSB?

I was trying the following way... I don't think that it's the right way:

value = 0x118;  

Storing in bytes...

result[5] = (byte) value;  
result[6] = (byte)(value << 8);  
...

What is the correct way?

Answer 1

This will do it:

result[5] = (byte) (value & 0xFF);           // Least significant "byte"
result[6] = (byte) ((value & 0xFF00) >> 8);  // Most significant "byte"

I usually use bit masks - maybe they're not needed. The first line selects the lower eight bits, the second line selects the upper eight bits and shifts the bits eight bit positions to the right. This is equal to a division by 2⁸.

---

This is the "trick" behind:

  (I) LSB

  01010101 10101010        // Input
& 00000000 11111111        // First mask, 0x00FF
  -----------------
  00000000 10101010        // Result - now cast to byte

  (II) MSB

  01010101 10101010        // Input
& 11111111 00000000        // Second mask, 0xFF00
  -----------------
  01010101 00000000        // Result - 
  >>>>>>>>                 // "Shift" operation, eight positions to the right
  -----------------
  00000000 01010101        // Result - now cast to byte

---

To sum it up, do the following calculation:

 byte msb = result[6];
 byte lsb = result[5];
 int result = (msb << 8) + lsb;    // Shift the MSB bits eight positions to the left.