java understanding bitwise manipulation

Question

Recently I have been learning about bitwise operators and along the way there is this code that finds the binary digits of a decimal number using the AND(&) bitwise operator, the code is as follows:

 byte b = -34;

 for(int t = 128;t > 0; t = t/2)
{
 if((b & t) != 0)System.out.println("1 ");
 else System.out.println("0 ");
 System.out.println("b & t yields: " + (b & t));
   }

I have modified the code to show the value calculated by b&t during each iteration. I would like to understand the exact mechanism behind this code as to why it works to find the binary digits, please explain why is b compared to t each iteration and why is t divided by 2 each iteration?

In addition, I would like to know how is (b&t) calculated manually by listing the binary digits.I do have an understanding how & works but when I listed out the binary digits of 34 and 128 and compared them:

1 0 0 0 0 0 0 0(128)
0 0 1 0 0 0 1 0(34) //I am unsure if the negative sign should be included
---------------
0 0 0 0 0 0 0 0

the result I got was 0, however the program returns 128 which is perplexing.

Below I will also include the result of the execution of the program:

1 
b & t yields: 128
1 
b & t yields: 64
0 
b & t yields: 0
1 
b & t yields: 16
1 
b & t yields: 8
1 
b & t yields: 4
1 
b & t yields: 2
0 
b & t yields: 0

Much obliged for the help :)

Nowadays computers use two's complement, so -34 is represented by 256-34 as a unsigned byte. — user202729
– user202729, Commented Jun 10, 2017 at 15:59
Thanks very much,it makes sense that -34&128 return 128 now :) — TheRelativisticPhiloSheep
– TheRelativisticPhiloSheep, Commented Jun 10, 2017 at 16:08

Hearen · Accepted Answer · 2019-04-02 01:22:29Z

2

Dividing t by 2 is a bit-shift to the right:

1 0 0 0 0 0 0 0 128 = t
0 1 0 0 0 0 0 0 64  = t / 2
0 0 1 0 0 0 0 0 32  = t / 2 / 2
...

t always has one bit set to 1, all others are 0.

Then you compare that to b using &. Each result bit is 1 if and only if the corresponding bit in both inputs is 1 as well. That means that we basically check if the bit in b is 1 at the location that the t-bit is 1. That is done for all bits from left to right.

edited Apr 2, 2019 at 1:22

Hearen

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answered Jun 10, 2017 at 16:09

luk2302

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Edward Aung · Accepted Answer · 2019-04-02 03:36:10Z

0

Since there were two questions: (1) OP is perplex why his hand-calculation does not match with program's code and (2) OP would like to know why one of the variables is divided by 2.

I merely combine the answers: (1) Negative numbers are represented as two's complement. Therefore negative 34 is

 0 0 1 0 0 0 1 0 <-- +34
 1 1 0 1 1 1 0 1 <-- one's complement of 34
 1 1 0 1 1 1 1 0 <-- two's complement of 34

Note: two's complement is one's complement + 1.

(2) Division by 2 is shifting to the right (if it is binary). That's why, the third time in the loop, it outputs zero (The only 1 in 128 is 'AND'ed with the third zero in -34).

answered Apr 2, 2019 at 3:36

Edward Aung

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