指向成員對象的指針 - 中斷線程

Question

這是對a question I got answered yesterday的跟進。

我試圖讓成員函數中斷線程，但我不知道是什麼，這些錯誤是：

[vagrant@localhost projects]$ g++ -o test test.cpp -lpthread 
test.cpp: In instantiation of ‘InterruptibleThread::InterruptibleThread(Function&&, Args&& ...)::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, Args&& ...)> [with auto:1 = void (MyClass::*)(int); Function = void (MyClass::*)(int); Args = {MyClass*, int}; std::atomic_bool = std::atomic<bool>]’: 
/usr/local/include/c++/6.3.0/type_traits:2481:26: required by substitution of ‘template<class _Fn, class ... _Args> static std::__result_of_success<decltype (declval<_Fn>()((declval<_Args>)()...)), std::__invoke_other> std::__result_of_other_impl::_S_test(int) [with _Fn = InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, MyClass*&&, int&&)>; _Args = {std::reference_wrapper<std::atomic<bool> >, std::reference_wrapper<std::atomic<bool> >, void (MyClass::*)(int), MyClass*, int}]’ 
/usr/local/include/c++/6.3.0/type_traits:2492:55: required from ‘struct std::__result_of_impl<false, false, InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, MyClass*&&, int&&)>, std::reference_wrapper<std::atomic<bool> >, std::reference_wrapper<std::atomic<bool> >, void (MyClass::*)(int), MyClass*, int>’ 
/usr/local/include/c++/6.3.0/type_traits:2496:12: required from ‘class std::result_of<InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, MyClass*&&, int&&)>(std::reference_wrapper<std::atomic<bool> >, std::reference_wrapper<std::atomic<bool> >, void (MyClass::*)(int), MyClass*, int)>’ 
/usr/local/include/c++/6.3.0/functional:1365:61: required from ‘struct std::_Bind_simple<InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, MyClass*&&, int&&)>(std::reference_wrapper<std::atomic<bool> >, std::reference_wrapper<std::atomic<bool> >, void (MyClass::*)(int), MyClass*, int)>’ 
/usr/local/include/c++/6.3.0/thread:137:26: required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]::<lambda(std::atomic_bool&, std::atomic_bool&, auto:1&&, MyClass*&&, int&&)>; _Args = {std::reference_wrapper<std::atomic<bool> >, std::reference_wrapper<std::atomic<bool> >, void (MyClass::*)(int), MyClass*, int}]’ 
test.cpp:41:33: required from ‘InterruptibleThread::InterruptibleThread(Function&&, Args&& ...) [with Function = void (MyClass::*)(int); Args = {MyClass*, int}]’ 
test.cpp:111:53: required from here 
test.cpp:36:9: error: must use ‘.*’ or ‘->*’ to call pointer-to-member function in ‘fxn (...)’, e.g. ‘(... ->* fxn) (...)’ 
     fxn(forward<Args>(args)...); 

當前代碼：

using namespace std; 

class InterruptThreadException {}; 

class InterruptibleThread { 
private: 
    static thread_local atomic_bool* stopRef; 
    static thread_local atomic_bool* pauseRef; 
    atomic_bool stopFlag{false}; 
    atomic_bool pauseFlag{false}; 
    thread thrd; 

public: 
    friend void checkForInterrupt(); 
    template < typename Function, typename... Args > 
    InterruptibleThread(Function&& _fxn, Args&&... _args) 
     : thrd(
       [](atomic_bool& sr, atomic_bool& pr, auto&& fxn, Args&&... args) { 
        stopRef = &sr; 
        pauseRef = &pr; 
        fxn(forward<Args>(args)...); 
       }, 
       ref(stopFlag), 
       ref(pauseFlag), 
       forward<Function>(_fxn), 
       forward<Args>(_args)...) { 
     thrd.detach(); 
    } 
    bool stopping() const { 
     return stopFlag.load(); 
    } 

    void stop() { 
     stopFlag.store(true); 
    } 

    void pause() { 
     pauseFlag.store(true); 
     cout << "setting pause flag: " << pauseFlag.load() << endl; 
    } 

    void start() { 
     pauseFlag.store(false); 
    } 
}; 

void checkForInterrupt() { 
    cout << "Pause flag: " << InterruptibleThread::pauseRef->load() << endl; 
    cout << "Stop flag: " << InterruptibleThread::stopRef->load() << endl; 
    while (InterruptibleThread::pauseRef->load()) { 
     cout << "Paused\n"; 
     this_thread::sleep_for(chrono::seconds(1)); 
    } 
    if (!InterruptibleThread::stopRef->load()) { 
     return; 
    } 
    throw InterruptThreadException(); 
} 

thread_local atomic_bool* InterruptibleThread::stopRef = nullptr; 
thread_local atomic_bool* InterruptibleThread::pauseRef = nullptr; 

void doWork() { 
    int i = 0; 
    try { 
     while (true) { 
      cout << "Checking for interrupt: " << i++ << endl; 
      checkForInterrupt(); 
      this_thread::sleep_for(chrono::seconds(1)); 
     } 
    } catch (InterruptThreadException) { 
     cout << "Interrupted\n\n"; 
    } 
} 

class MyClass { 
private: 
    int myInt; 
    void setInt(int i) { 
     myInt = i; 
    } 

public: 
    MyClass() : myInt(1) { 
    } 
    void myWork(int i); 
    void doWork(); 
}; 

void MyClass::myWork(int i) { 
    setInt(i); 
    cout << "myInt value: " << myInt << endl; 
} 

void MyClass::doWork() { 
    InterruptibleThread t(&MyClass::myWork, this, 666); 
} 

int main() { 
    MyClass mc; 
    mc.doWork(); 

    cout << "Press enter to exit" << endl; 
    getchar(); 
    return 0; 
} 

我試過編譯器的建議，得到了之後的錯誤與fold表達式有關（afaik是C++ 17，我不會使用超過14的任何東西）。任何想法如何讓這個工作？

我有一個沒有使用lambdas的工作版本，但我真的好奇如何讓這與原始代碼一起工作。

Answer 1

指向成員函數的指針使用與指向非成員函數或其他函數的指針不同的語法來調用。由於您的案例中的Function是void (MyClass::*)(int)，因此您需要使用語法(object.*fxn)(arg)或(objectPtr->*fxn)(arg)來調用它。

如果你有訪問C++ 17層的功能，你可以使用std::invoke均勻撥打不同類型的可調用的：

InterruptibleThread(Function&& _fxn, Args&&... _args) 
    : thrd(
      [](atomic_bool& sr, atomic_bool& pr, auto&& fxn, auto&&... args) { 
       stopRef = &sr; 
       pauseRef = ≺ 
       std::invoke(std::move(fxn), std::move(args)...); 
      }, 
      //... 

注：我改變Args&&... args到auto&&... args因爲Args&&...可能會導致問題。

如果您沒有訪問C++ 17，你可以自己實現invoke相當容易：

template <typename Callable, 
      typename... Args, 
      typename = std::enable_if_t<!std::is_member_function_pointer<Callable>::value>> 
decltype(auto) invoke(Callable&& c, Args&&... args) { 
    return std::forward<Callable>(c)(std::forward<Args>(args)...); 
} 

template <typename T, 
      typename T2, 
      typename Ret, 
      typename... FuncArgs, 
      typename... Args, 
      typename = std::enable_if_t<!std::is_pointer<T2>::value>> 
decltype(auto) invoke(Ret (T::*c)(FuncArgs...), T2&& t, Args&&... args) { 
    return (std::forward<T>(t).*c)(std::forward<Args>(args)...); 
} 

template <typename T, 
      typename T2, 
      typename Ret, 
      typename... FuncArgs, 
      typename... Args> 
decltype(auto) invoke(Ret (T::*c)(FuncArgs...), T2* t, Args&&... args) { 
    return (t->*c)(std::forward<Args>(args)...); 
} 

這並不盡一切std::invoke做，但它的一切，你需要它。