C++ 11線程初始化與成員函數編譯錯誤

Question

我剛剛開始使用C++ 11線程，我一直在努力（可能很愚蠢）的錯誤。 這是我的示例程序：

#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; 
} 

是否有可能從一個純粹的成員函數啓動一個線程？如果不是，我怎麼能包裝我的foo函數從對象obj能夠創建這樣的線程？ 在此先感謝！

這是編譯錯誤：

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

你需要調用對象採取任何參數，因此

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

應該做的伎倆。這具有創建可調用實體的效果，其在obj上調用A::foo。

原因是A的非靜態成員函數採用類型的隱式第一個參數（可能是cv限定的）A*。當您致電obj.foo()時，您實際上呼叫A::foo(&obj)。一旦你知道了，上面的咒語是非常有道理的。