C++ 11线程初始化与成员函数编译错误

Question

我刚刚开始使用C++ 11线程，我一直在努力（可能很愚蠢）的错误。 这是我的示例程序：

#include <iostream> 
#include <thread> 
#include <future> 
using namespace std; 

class A { 
public: 
    A() { 
    cout << "A constructor\n"; 
    } 

    void foo() { 
    cout << "I'm foo() and I greet you.\n"; 
    } 

    static void foo2() { 
    cout << "I'm foo2() and I am static!\n"; 
    } 

    void operator()() { 
    cout << "I'm the operator(). Hi there!\n"; 
    } 
}; 

void hello1() { 
    cout << "Hello from outside class A\n"; 
} 

int main() { 
    A obj; 
    thread t1(hello1); // it works 
    thread t2(A::foo2); // it works 
    thread t3(obj.foo); // error 
    thread t4(obj);  // it works 

    t1.join(); 
    t2.join(); 
    t3.join(); 
    t4.join(); 
    return 0; 
} 

是否有可能从一个纯粹的成员函数启动一个线程？如果不是，我怎么能包装我的foo函数从对象obj能够创建这样的线程？ 在此先感谢！

这是编译错误：

thread_test.cpp: In function ‘int main()’: thread_test.cpp:32:22: error: no matching function for call to ‘std::thread::thread()’

thread_test.cpp:32:22: note: candidates are:

/usr/include/c++/4.6/thread:133:7: note: std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (A::*)(), _Args = {}]

/usr/include/c++/4.6/thread:133:7: note: no known conversion for argument 1 from ‘’ to ‘void (A::*&&)()’

/usr/include/c++/4.6/thread:128:5: note: std::thread::thread(std::thread&&)

/usr/include/c++/4.6/thread:128:5: note: no known conversion for argument 1 from ‘’ to ‘std::thread&&’

/usr/include/c++/4.6/thread:124:5: note: std::thread::thread()

/usr/include/c++/4.6/thread:124:5: note: candidate expects 0 arguments, 1 provided

Answer 1

你需要调用对象采取任何参数，因此

thread t3(&A::foo, &obj); 

应该做的伎俩。这具有创建可调用实体的效果，其在obj上调用A::foo。

原因是A的非静态成员函数采用类型的隐式第一个参数（可能是cv限定的）A*。当您致电obj.foo()时，您实际上呼叫A::foo(&obj)。一旦你知道了，上面的咒语是非常有道理的。