Bit computing common finishing operations (C ++ to achieve)

Number 1

Solution 1:

With the n & 1rightmost determines whether a 1, then continues to determine the right the rightmost one, is repeated until n = 0. The time complexity isO(log2n)

#include <iostream>
using namespace std;
int main() {
    int n = 0b1101, cnt = 0;
    while (n) {
        cnt += n & 1;
        n >>= 1;
    }
    cout << cnt << endl;
    return 0;
}

Solution 2:

With n & (n - 1)erasable rightmost 1 is repeated until n = 0. The time complexity is O(count), depending on the number of count 1, this method is better than the first.

#include <iostream>
using namespace std;
int main() {
    int n = 0b1101, cnt = 0;
    while (n) {
        n = n & (n - 1);
        ++cnt;
    }
    cout << cnt << endl;
    return 0;
}

For determining whether a bit is 1 or 0

Analyzing the k-th position from the right,(n >> (k - 1)) & 1

#include <iostream>
using namespace std;
int main() {
    int n = 0b1101;
    cout << (n >> 1 & 1) << endl;  // 第 2 位为 0
    cout << (n >> 2 & 1) << endl;  // 第 3 位为 1
    return 0;
}

Reversed a bit

Reversal from the right k-th position,n ^ (1 << (k - 1))

#include <iostream>
using namespace std;
int main() {
    int n = 0b1101;
    cout << (n ^ (1 << 2)) << endl;  // 反转第 3 位 -> 1001
    cout << (n ^ (1 << 1)) << endl;  // 反转第 2 位 -> 1111
    return 0;
}

Some practical rules

Bitwise XOR a same number twice ( n ^ x ^ x), unchanged.

#include <iostream>
using namespace std;
int main() {
    int n = 13, x = 666;
    cout << (n ^ x ^ x) << endl;  // 仍然是 13
    return 0;
}

Bitwise with the mold in place, parity determination, more efficient. n & 1The result is an even number is 0, the result is an odd number is 1.

#include <iostream>
using namespace std;
int main() {
    cout << (4 & 1) << endl;
    cout << (5 & 1) << endl;
    return 0;
}

K k bits to the left take the power of 2, k bits to the right by 2 k-th power. It is common in the divide and conquer strategy, such as merge sort:

#include <cstdlib>
#include <ctime>
#include <iostream>
using namespace std;

void partition(int a[], int l, int m, int r) {
    int *t = new int[r - l + 1];
    int i = l, j = m + 1, k = 0;
    while (i <= m && j <= r) t[k++] = a[i] < a[j] ? a[i++] : a[j++];
    while (i <= m) t[k++] = a[i++];
    while (j <= r) t[k++] = a[j++];
    for (int i = l; i <= r; ++i) a[i] = t[i - l];
    delete[] t;
}

void merge_sort(int a[], int l, int r) {
    if (l < r) {
        int m = (l + r) >> 1;  // 相当于 (l + r) / 2
        merge_sort(a, l, m);
        merge_sort(a, m + 1, r);
        partition(a, l, m, r);
    }
}

int main() {
    srand(time(0));
    int a[20];
    for (int i = 0; i < 20; ++i) a[i] = rand() % 100;
    merge_sort(a, 0, 19);
    for (int i = 0; i < 20; ++i) cout << a[i] << " ";
    return 0;
}