[Sword refers to offer] The number of 1 in binary

Topic description:

Input an integer and output the number of 1's in the binary representation of the number. Negative numbers are represented in complement.

Problem solving plan:

for(int i=0; i<32; i++){
	t = (n & 0x80000000 >>> i) >>> (31-i);
	if(t > 0)
	     totalOf1 ++;
  }

Split this code. To understand this shift output problem, you need to understand the following:
* 0x80000000 is the hexadecimal representation of the number, and the binary representation is 1000000000000000000000000000000
* The priority of the operation, the shift operation is higher than Logical operation, >>> higher than &
* bit logical AND operation 1&1 = 1, 0&1 = 0
* >>> unsigned right shift, part of the shift is discarded, and the left bit is filled with 0;

the execution sequence of the statement block of the for loop:
1. 0x80000000 unsigned right shift i bits;
2, n and 1 result bitwise AND;
3, 2 result unsigned right shift 31-i bits
4, output 3 result

The difference between >> and >>>:

>>: Signed right shift . A positive number is shifted to the right and the high-order position is filled with 0, and a negative number is shifted to the right and the high-order position is filled with 1. for example:

4 >> 1, the result is 2; -4 >> 1, the result is -2. -2 >> 1, the result is -1.
>>>: Unsigned right shift . Whether it is a positive or negative number, the high bits are filled with 0.
For positive numbers, >> and >>> are no different.
For negative numbers, -2 >>> 1, the result is 2147483647 (Integer.MAX_VALUE), and -1 >>> 1, the result is 2147483647 (Integer.MAX_VALUE).

Code:

import java.util.Scanner;

public class NumberOf1 {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		while(in.hasNext()){
			int n = in.nextInt();
			int result = numberOf1(n);
			System.out.println(result);
		}
	}
	
	public static int numberOf1(int n){
		int t = 0;
		int totalOf1 = 0;
		for(int i=0; i<32; i++){
			t = (n & 0x80000000 >>> i) >>> (31-i);
			if(t > 0)
				totalOf1 ++;
		}
		return totalOf1;
	}
}