JAVA Bitwise code purpose , &

Question

// following code prints out Letters aA bB cC dD eE ....

class UpCase {
public static void main(String args[]) {
 char ch;

 for(int i = 0; i < 10; i++) {
  ch = (ch         


        

Answer 1

UpCase

The decimal number 66503 represented by a 32 bit signed integer is 00000000 00000001 00000011 11000111 in binary.

The ASCII letter a represented by a 8 bit char is 01100001 in binary (97 in decimal).

Casting the char to a 32 bit signed integer gives 00000000 00000000 00000000 01100001.

&ing the two integers together gives:

00000000 00000000 00000000 01100001
00000000 00000001 00000011 11000111
===================================
00000000 00000000 00000000 01000001

which casted back to char gives 01000001, which is decimal 65, which is the ASCII letter A.

Showbits

No idea why you think that 128, 64 and 32 are all 10000000. They obviously can't be the same number, since they are, well, different numbers. 10000000 is 128 in decimal.

What the for loop does is start at 128 and go through every consecutive next smallest power of 2: 64, 32, 16, 8, 4, 2 and 1.

These are the following binary numbers:

128: 10000000
 64: 01000000
 32: 00100000
 16: 00010000
  8: 00001000
  4: 00000100
  2: 00000010
  1: 00000001

So in each loop it &s the given value together with each of these numbers, printing "0 " when the result is 0, and "1 " otherwise.

Example:

val is 123, which is 01111011.

So the loop will look like this:

128: 10000000 & 01111011 = 00000000 -> prints "0 "
 64: 01000000 & 01111011 = 01000000 -> prints "1 "
 32: 00100000 & 01111011 = 00100000 -> prints "1 "
 16: 00010000 & 01111011 = 00010000 -> prints "1 "
  8: 00001000 & 01111011 = 00001000 -> prints "1 "
  4: 00000100 & 01111011 = 00000000 -> prints "0 "
  2: 00000010 & 01111011 = 00000010 -> prints "1 "
  1: 00000001 & 01111011 = 00000001 -> prints "1 "

Thus the final output is "0 1 1 1 1 0 1 1", which is exactly right.