CODEWITHSUNDEEP
javabit-manipulation
Umesh Awasthi
Umesh Awasthi

Reputation: 23587

bitwise right shift and 0xFF | Java

I am trying to understand a piece of code but not able to get clear idea about few points here is the Java code

private String firstMethod(int number){
   return secondMethod(number >> 16 & 0xFF, 0).concat(secondMethod(number >> 8 & 0xFF, 1)).concat(secondMethod(number & 0xFF, 7));
}

private static String secondMethod(int value, int offset)
 {
     return thirdMethod(value >> 4, offset).concat(thirdMethod(value & 0xF, offset + 4));
 }

private static String thirdMethod(int value, int offset)
  {
     String chars = getAlphabet();
     int pos = (value + offset) % 16;
     return chars.substring(pos, pos + 1);
 }

value passed to firstMethod is a random number for first time and all subsequent call to method will pass value incremented by 1.

I am clear about bit-wise right shift operation as well about the use of & 0xFF, however I am still not very clear about following points

Can anyone help me to understand those 2 point

Upvotes: 3

Views: 1341

Answers (2)

Ecki
Ecki

Reputation: 11

An integer is a 32-Bit Number.

So as a binary-number, you can represent number as:

XXXXXXXXAAAAAAAABBBBBBBBCCCCCCCC (X, A, B, C stands for 0 or 1).

number >> 16 gives you XXXXXXXXAAAAAAAA. number >> 16 & 0xFF gives you AAAAAAAA

By the firstMethod number is splited in 3 Bytes:

AAAAAAAA and BBBBBBBB and CCCCCCC (Shift of 16, shift of 8 and no shift) and given to the secondMethod.

In the secondMethod the 8 Bits are splited in the higher four bits and the lower four bits.

In the thirdMethod the four Bits is translate to a String containing one char.

But the sense depends on "getAlphabet()". Perhaps there will be also a usefull interpretation for the offset.

So you have to give further information!

Upvotes: 1

Peter Lawrey
Peter Lawrey

Reputation: 533530

Shifting given value by specific number (like 16 and 8 for first than no sift etc)

You are printing bytes as hexi-decimal. Each byte is 8-bits so you want to shift each byte by

Not clear about use of offset ,specifically passing certain number as offset.

I am pretty sure the offset is either a) incorrect, b) a really obscure way of masking/encoding the data.

To print a number as a 6 byte hexi-decimal String you can do this.

System.out.println(String.format("%06x", 12345));

prints

003039

This is much shorter. ;)

>> has a surprising low precedence. This means

number >> 16 & 0xFF

is actually

number >> (16 & 0xFF)

or

number >> 16

what you indedn was

(number >> 16) & 0xFF

or as the result is unsigned.

(number >>> 16) & 0xFF

Upvotes: 3

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