c++ - 使用构造函数参数将仿函数传递给 std::thread。是否可以？

Question

class worker {
    std::atomic_bool done;
    std::string segment_name;
    bi::named_mutex mutex;
    std::string file_name;
    data_segment_state state;
    unsigned long data_size;
    std::unique_ptr<unsigned char> data;
protected:
    void operator()() {
        while(!done) {
        }
    }

public:
    worker(const std::string& _segment_name, bi::managed_shared_memory& segment)
    : done{false},
          segment_name{_segment_name},
          mutex{bi::open_or_create, segment_name.c_str() },
          file_name {},
          state {data_segment_state::not_received },
          data_size{0},
          data {segment.construct<unsigned char>(segment_name.c_str())[chunk_size](0) } 
    }

    worker(worker&& rhs)
    : done {rhs.done.load()} ,
      mutex(bi::open_or_create, rhs.segment_name.c_str()),
      segment_name{rhs.segment_name},
      file_name {rhs.file_name},
      state {rhs.state },
      data_size{rhs.data_size},
      data {std::move(rhs.data)} {
    }
};

...

 std::string worker_name;
        std::thread t{worker{worker_name, segment}};

In instantiation of 'struct std::_Bind_simple<worker()>':
thread:137:47:   required from 'std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = worker; _Args = {}]'
error: no type named 'type' in 'class std::result_of<worker()>'
       typedef typename result_of<_Callable(_Args...)>::type result_type;

什么是函数对象的有效线程初始化？错误全文: 错误:“class
”中没有名为“type”的类型 std::result_of' typedef typename result_of<_Callable(_Args...)>::type result_type; ^

Answer 1

只需传递构造的仿函数:

std::thread t(worker{str, 10});

Live demo

constructor of std::thread您指的是采用构造的函数对象和调用该函数对象时将传递的参数。因此，在您的情况下，参数将传递给 operator()。