Common bits operations in Java

One, calculates the number of 1 in the binary representation of a positive

1     //number of digits in binary notation for solving positive numbers 2     private static int countbit (int num) {3         int count = 0;4 for         (; num > 0; count++) 5         {6             num &= (num-1); 7         }8         return count;9     }

Algorithm idea: Every for loop, the rightmost 1 of NUM's binary is cleared.

Why is n &= (n–1) able to erase the rightmost 1? Because from the binary point of view, N is equivalent to the lowest bit of n-1 plus 1. For example, 8 (1000) = 7 (0111) + 1 (0001), so 8 & 7 = (+) & (0111) = 0 (0000), the 8 rightmost 1 is cleared (actually the highest 1, because there is only one 8 in the binary of 1). For example, 7 (0111) = 6 (0110) + 1 (0001), so 7 & 6 = (0111) & (0110) = 6 (0110), the rightmost 7 of the binary representation of 1 (that is, the lowest bit 1) is cleared.

Reference: algorithm-Finding the number of 1 in a binary number

Two, get the first bit of a number (to determine if the first bit of a number is 0 or 1?). )Train of thought: if the first and 1 phase and the result of 1 indicate that the first bit is 1; If 0 indicates that the I bit is 0

    Gets the value of the I-bit of the integer num    private static Boolean getbit (int num, int i)    {        return (num & (1 << i)) = 0); True indicates that the first bit is 1, otherwise 0    }

1 left I bit, get a number, this number is only the first I bit 1, the other bits are 0

Num with this number, the result is either 0, or not 0. The result is a non-0 indicating that the I bit is 1, and the result is 0 that the I bit is 0

Third, set the I bit to 1Idea: The first bit with 0 or, the value does not change. The first bit with 1 or, becomes 1. Therefore, our goal is num with an i-bit value of 1, the value of the other bits is 0 of the number of phases or

    Place the value of the I bit of the integer num to 1    private static int getbit (int num, int i)    {        return (num | (1 << i));    }

Four, set the I bit to 0 (clear 0)Idea: The first and 0 and the first I bit will become 0. The other bits are the same as the 1 and the other bits remain unchanged. In this way, I'm only going to clear the first 0.

    Place the value of the I bit of the integer num to 1    private static int getbit (int num, int i)    {       int mask = ~ (1 << i);//000100
       Return (num & mask);//111011    }

In short, to get the value of the I-bit, or I position 0 or 1, the overall idea is: Move 1 left I bit, and then proceed with operation or OR operation.

Transferred from: http://www.cnblogs.com/hapjin/p/5839797.html

Common bits operations in Java (RPM)