有人可以向我解释这个按位程序吗？

Question

public class UpCase {
    public static void main(String[] args) {
        int t;
        byte val;
        val = 123;
        for (t = 128; t > 0; t = t / 2) {
            System.out.println(t);
            if ((val & t) != 0) System.out.println(" 1");
            else System.out.println(" 0");
        }
    }
}

In particular, I am not sure why we are using val=123 ? I understand that this program will print out 123 in binary but why is that the case? How does this work? I do understand however, the principles of the & operator and how you can turn on and off bits but I am not sure how it works in this particular example?

Answer 1

This program will print out the binary digits of the number in val from MSB to LSB by comparing it to each power of 2:

123 : 01111011 &
128 : 10000000 = 
      00000000 

      00000000 != 0 => false, print 0

123 : 01111011 &
 64 : 01000000 = 
      01000000 

      01000000 != 0 =>  true, print 1

123 : 01111011 &
 32 : 00100000 = 
      00100000 

      00100000 != 0 =>  true, print 1

// repeat for 2^4-2^1... 

123 : 01111011 &
  1 : 00000001 = 
      00000001 

      00000001 != 0 =>  true, print 1

Answer 2

Very simple:

It just checks if the value (123 in this case) using the bitwise operator &. The result of that is 1 or 0, this process is repeated for the following value 0.5t etc until t=0, resulting in the binary string for this value 123.