c++ - 创建线程时出现 C++ 错误，静态断言失败：转换为右值后，std::thread 参数必须是可调用的

Question

错误信息如下

/usr/include/c++/9/thread: In instantiation of ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(DisplayObject&); _Args = {DisplayObject&}; <template-parameter-1-3> = void]’:
farmville.cpp:242:53:   required from here
/usr/include/c++/9/thread:120:44: error: static assertion failed: std::thread arguments must be invocable after conversion to rvalues
  120 |           typename decay<_Args>::type...>::value,
      |                                            ^~~~~
/usr/include/c++/9/thread: In instantiation of ‘struct std::thread::_Invoker<std::tuple<void (*)(DisplayObject&), DisplayObject> >’:
/usr/include/c++/9/thread:131:22:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(DisplayObject&); _Args = {DisplayObject&}; <template-parameter-1-3> = void]’
farmville.cpp:242:53:   required from here
/usr/include/c++/9/thread:243:4: error: no type named ‘type’ in ‘struct std::thread::_Invoker<std::tuple<void (*)(DisplayObject&), DisplayObject> >::__result<std::tuple<void (*)(DisplayObject&), DisplayObject> >’
  243 |    _M_invoke(_Index_tuple<_Ind...>)
      |    ^~~~~~~~~
/usr/include/c++/9/thread:247:2: error: no type named ‘type’ in ‘struct std::thread::_Invoker<std::tuple<void (*)(DisplayObject&), DisplayObject> >::__result<std::tuple<void (*)(DisplayObject&), DisplayObject> >’
  247 |  operator()()

我要做的是创建一个控制对象的线程。所以我创建了一个可以控制对象的函数，并使用该函数创建一个线程。

void chickenAction(DisplayObject &chicken){
    int x = std::max(1, chicken.current_x + (1 + std::rand()) % 10 - 5);
    int y = std::max(1, chicken.current_y + (1 + std::rand()) % 10 - 5);
    chicken.draw(y, x);
    chicken.moveto(y, x);
}

下面是创建线程的代码，chicken1 是一个对象：

std::thread chicken1Thread(chickenAction, chicken1);
chicken1Thread.join();

我认为下面的代码可以重复错误

#include <iostream>
#include <unistd.h>
#include <thread>

class DisplayThread {
 public:
  int size;
  DisplayThread(){};
  void changeSize(int x) {
      size = x;
  }
};

void displayAction(DisplayThread &td){
    td.changeSize(1);
}



int main() {
        DisplayThread td;
        std::thread displaytd(displayAction, td);
    displaytd.join();
    return 0;
}

标签： c++multithreadingc++11concurrencyc++17

Answer 1

lambda 签名需要一个引用。但是，std::thread不能接受引用：它必须能够将对象复制到它调用的函数中。

解决方案是std::reference_wrapper及其方便的小包装器std::ref和std::cref.

这是一个工作示例：

#include <thread> // std::thread
#include <functional> // std::ref

void foo() {
    int i_ = 0;
    auto thread_func = [](int &i) {
        return i+42;
    };

    // assertion failed: std::thread arguments must be invocable after conversion to rvalues
    // std::thread t{thread_func, i_};

    // OK!
    std::thread t{thread_func, std::ref(i_)};
    t.join();
}