将监视器方法作为线程参数c++[duplicate]传递

list<int> l;
class monitor{
    public:
    void m_add(int x){
        //lock use of m_remove
        //lock use of m_add
        l.push_back(x);
        //unlock use of m_add
        //unlock use of m_remove
    }
    void m_remove(int x){
        //lock use of m_remove
        //lock use of m_contains
        //lock use of m_add
        l.remove(x);
        //unlock use of m_add
        //unlock use of m_contains
        //unlock use of m_remove
    }
    bool m_contains(int x){
        //lock use of m_remove
        bool found = find(l.begin(), l.end(), x) != l.end();
        //unlock use of m_remove
        return found;
    }
    private:
    mutex mtx_remove;
    mutex mtx_add;
    mutex mtx_contains;
};

在main函数中,我创建了一个线程来运行add函数,例如。 thread t(m.m_add, 1); 我得到了错误

error: invalid use of non-static member function

我理解(通过查看这里的其他答案),我应该像中那样运行线程 thread t(&monitor::m_add, 1); 并将所有方法声明为静态,但我需要实例化 monitor 为了创建互斥锁(或锁,或其他什么),并使它们在main的作用域中是私有的。

对于这种情况,最合适的解决方案是什么?如果锁只在monitor类的范围内可访问,那将很有趣(顺便说一句,我还打算将 list<int> l 以后在monitor类中。)