C++中实现位运算的示例代码

#include <iostream>
#include <iomanip>  // 仅用于格式化二进制输出，无额外子函数
using namespace std;

int main() {
    
    cout << "=== 基础位运算示例 ===" << endl;
    int a = 6;   // 二进制：00000110
    int b = 10;  // 二进制：00001010

    // 1. 按位与 &
    int andResult = a & b;  // 00000010 → 2
    cout << "a & b = " << andResult << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << ((andResult >> i) & 1);
    cout << endl;

    // 2. 按位或 |
    int orResult = a | b;   // 00001110 → 14
    cout << "a | b = " << orResult << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << ((orResult >> i) & 1);
    cout << endl;

    // 3. 按位异或 ^
    int xorResult = a ^ b;  // 00001100 → 12
    cout << "a ^ b = " << xorResult << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << ((xorResult >> i) & 1);
    cout << endl;

    // 4. 按位取反 ~（仅显示低8位）
    int notResult = ~a;
    cout << "~a = " << notResult << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << (((unsigned int)notResult & 0xFF) >> i & 1);
    cout << endl;

    // 5. 左移 <<
    int leftShift = a << 2; // 00011000 → 24
    cout << "a << 2 = " << leftShift << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << ((leftShift >> i) & 1);
    cout << endl;

    // 6. 右移 >>
    int rightShift = b >> 1; // 00000101 → 5
    cout << "b >> 1 = " << rightShift << "  ";
    for (int i = 7; i >= 0; i--) 
        cout << ((rightShift >> i) & 1);
    
    cout << "\n=====================" << endl;

    // ===================== 第二部分：位运算实用技巧 =====================
 
    // 技巧1：判断奇偶
    int num1 = 15, num2 = 16;
    cout << num1 << " 是" << ((num1 & 1) ? "奇数" : "偶数") << endl;
    cout << num2 << " 是" << ((num2 & 1) ? "奇数" : "偶数") << endl;

    // 技巧2：交换两个数（无需临时变量）
    int x = 8, y = 12;
    cout << "\n交换前：x=" << x << ", y=" << y << endl;
    x = x ^ y;
    y = x ^ y;
    x = x ^ y;
    cout << "交换后：x=" << x << ", y=" << y << endl;

    // 技巧3：将某一位置1（第3位，从0计数）
    int val = 5; // 00000101
    int bitPos = 3;
    val |= (1 << bitPos); // 00001101 → 13
    cout << "\n将第" << bitPos << "位置1后：" << val << endl;

    // 技巧4：将某一位清零（第2位）
    val = 13; // 00001101
    bitPos = 2;
    val &= ~(1 << bitPos); // 00001001 → 9
    cout << "将第" << bitPos << "位清零后：" << val << endl;

    // 技巧5：判断某一位是否为1（第1位）
    val = 9; // 00001001
    bitPos = 1;
    bool is1 = (val & (1 << bitPos)) != 0;
    cout << "第" << bitPos << "位是" << (is1 ? "1" : "0") << endl;

    // 技巧6：快速计算2的n次方
    bitPos = 5;
    int pow2 = 1 << bitPos; // 32
    cout << "2^" << bitPos << " = " << pow2 << endl;

    
    return 0;
}