operator[]();
operator<() ;
operator<=();
operator>();
operator>=();
operator&=();
operator|=();
operator^=();
operator&();
operator|();
operator[]();
这里有一道题
目的是写出类里面的函数。

#include<iostream>

using namespace std;
int main() {
    bitset a, b;
    int n, m, q;
    cin >> n >> m >> q;
    for (int i = 0; i < n; i++) {
        int x;
        cin >> x;
        a.set(x);
    }
    cout << "a.count() is " << a.count() << "\n";
    cout << "a.test(5) is " << (a.test(5) ? "true" : "false") << "\n";
    cout << "a.any() is " << (a.any() ? "true" : "false") << "\n";
    cout << "a.none() is " << (a.none() ? "true" : "false") << "\n";
    cout << "a.all() is " << (a.all() ? "true" : "false") << "\n";
    b = ~b;
    for (int i = 0; i < m; i++) {
        int x;
        cin >> x;
        b.reset(x);
    }
    cout << a << "\n";
    cout << b << "\n";
    if (a == b) {
        cout << "hello\n";
    }
    if (a != b) {
        cout << "world\n";
    }
    bitset c;
    // test &
    c = a;
    c &= b;
    cout << c << "\n";
    c = a & b;
    cout << c << "\n";
    // test |
    c = a;
    c |= b;
    cout << c << "\n";
    c = a | b;
    cout << c << "\n";
    // test ^
    c = a;
    c ^= b;
    cout << c << "\n";
    c = a ^ b;
    cout << c << "\n";
    // test <<
    c = a;
    c <<= 2;
    cout << c << "\n";
    c = a << 2;
    cout << c << "\n";
    // test >>
    c = b;
    c >>= 2;
    cout << c << "\n";
    c = b >> 2;
    cout << c << "\n";
    // test []
    for (int i = 0; i < q; i++) {
        int x;
        cin >> x;
        if (a[i])
            cout << "Yes\n";
        else
            cout << "No\n";
    }
}
#define N 5
const int max_length = 32 * N;//这个数字共有32*5位。从第0位到第max_length-1位，第0位是最低位。
class bitset {
private:
int a[N];//一个int32位，所以只用开N个int
public:
bitset();//默认构造函数，所有位初始化为0
void set(int pos);//把位置pos设置成1
void reset(int pos);//将位置pos设置成0
int count() const;//输出一共有多少个为1的位
bool test(int pos) const;//位置pos是否是1
bool any() const;//是否有是1的位
bool none() const;//是否没有是1的位
bool all() const;//是否所有位都是1

//位运算部分和普通数字的位运算相同。
bitset& operator&= (const bitset& b);
bitset& operator|= (const bitset& b);
bitset& operator^= (const bitset& b);
bitset& operator= (const bitset& b);
bitset& operator <<= (int pos);
bitset& operator >>= (int pos);
bitset operator~() const;
bitset operator&(const bitset& b) const;
bitset operator|(const bitset& b) const;
bitset operator^(const bitset& b) const;
bitset operator<<(int pos) const;
bitset operator>>(int pos) const;
bool operator== (const bitset& b) const;
bool operator!= (const bitset& b) const;
bool operator[] (int pos) const;//返回位置pos是不是1.

除了operator[]()，其它的我在网上找到了很有用的东西，感兴趣的可以去看一下。
http://