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我创建了两个类cl1，cl2并且cl1有一个带cl2&参数的构造函数。我有三个函数，一个cl1作为参数，一个cl1&&作为参数，一个cl1&作为参数。

#include <thread>
#include <iostream>

class cl1;
class cl2;


class cl2 {
public:
    int y;
    cl2(int y) : y(y)   {}                  //ctor
};

class cl1 {
public:
    int x;
    cl1(int x) : x(x) {}                 //ctor
    cl1(cl2& ob1) : x(ob1.y * 2) {}      //ctor for automatic conversion of cl2& to cl1, x = y*2
};

void do_work_with_cl(cl1 ob) {              //This works as usual by actually copying the object through the conversion constructor
    std::cout << "The x of ob is " << ob.x << std::endl;
}

void do_work_with_cl_rref(cl1&& ob) {       //I guess this works because it takes an rvalue and the automatic
                                            //conversion ctor of cl1 does just that
    std::cout <<"Inside the function that takes cl1 as rvalue, x of ob is"  << ob.x << std::endl;
}

void do_work_with_cl_lref(cl1& ob) {        //This doesn't work as ob is non-const lvalue reference
    std::cout << "lvalue referenced but the object created through implicit conversion is temporary(i.e rvalue)" << std::endl;
}   


int main() {
    //Normal non-threaded calls
    cl2 ob(100);                //create a cl2 object
    do_work_with_cl(ob);            //This is ok
    do_work_with_cl_rref(ob);   //This too works
    //do_work_with_cl_lref(ob)  //This fails, as suspected

    std::cout << "Thread part" << std::endl

    //Now calling the functions through a thread
    std::thread t1(do_work_with_cl_rref, ob);   //Thought this could work here, but doesn't
                                                //The other functions also don't work, but I can understand why.
    t1.join();                                              
}

在 ideone.com ：http://ideone.com/MPZc4C，因为我要问这个问题，所以这个例子有效。但是使用 g++-4.7 我收到如下错误：

In file included from /usr/include/c++/4.7/ratio:38:0,
             from /usr/include/c++/4.7/chrono:38,
             from /usr/include/c++/4.7/thread:38,
             from main.cpp:1:
/usr/include/c++/4.7/type_traits: In instantiation of ‘struct std::_Result_of_impl<false, false, void (*)(cl1&&), cl2>’:
/usr/include/c++/4.7/type_traits:1857:12:   required from ‘class std::result_of<void (*(cl2))(cl1&&)>’
/usr/include/c++/4.7/functional:1563:61:   required from ‘struct std::_Bind_simple<void (*(cl2))(cl1&&)>’
/usr/include/c++/4.7/thread:133:9:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(cl1&&); _Args = {cl2&}]’
main.cpp:13:44:   required from here
/usr/include/c++/4.7/type_traits:1834:9: error: invalid initialization of reference of type     ‘cl1&&’ from expression of type ‘cl2’
make: *** [main.o] Error 1

我真的不知道实现或代码是否有任何问题。我只是在学习 C++ 中的线程和东西，所以我这样做没有实际理由。请让我知道问题是什么，以及我在代码注释中是否正确。（代码中的注释“This works...”意味着当从 main() 将对象作为参数（而不是对它的引用）调用它们时它们很好。）

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1 回答 1

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C++ 标准的第 30.3.1.2/3 段说：

“要求：F 和 Args 中的每个 Ti 应满足 MoveConstructible 要求。INVOKE ( DECAY_COPY(std::forward<F>(f)), DECAY_COPY(std::forward<Args>(args))...)(20.8.2) 应为有效表达式”。

该表达式DECAY_COPY(x)又在 30.2.6 中定义：

“在本条款的几个地方DECAY_COPY(x)使用了操作。所有这些使用都意味着调用函数decay_copy(x)并使用结果，其中decay_copy定义如下：”

template <class T> typename decay<T>::type decay_copy(T&& v)
{ return std::forward<T>(v); }

由于该decay操作从对象中删除了 cv 限定符，因此需要有一个普遍有效的转换构造函数或从 typecl1到 type的转换运算符cl2。为了检查这一点，转发机制std::thread显然会生成右值引用cl1并尝试从中获取实例c2。这会失败，因为右值引用不能绑定到转换构造函数中的非常量左值引用。

如果您将构造函数的签名从更改cl1(cl2& ob1)为cl1(cl2 const& ob1)它适用于 GCC 4.7.2，因为右值引用可以绑定到对const.

于 2013-01-16T18:56:17.917 回答