关于std::thread中的c++自动类型转换行为

Regarding C++ automatic type conversion behaviour in std::thread

本文关键字：类型 转换 换行 c++ std thread 中的 关于      更新时间：2023-10-16

我创建了两个类cl1和cl2, cl1有一个构造函数，它接受cl2&参数。我有三个函数，一个以cl1为参数，一个以cl1&&为参数，一个以cl1&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，当我要问这个问题时，这个例子是有效的。但是在g++-4.7中，我得到了如下错误:

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

我真的不知道它是否有任何问题与实现，或代码…我只是在学习c++中的线程和东西，所以我这样做没有实际的原因。请让我知道问题是什么，如果我在代码的评论是正确的。(代码中的注释"This works…"意味着它们在以对象作为参数(而不是对它的引用)从main()调用时是好的)

c++标准第30.3.1.2/3段规定:

"要求:F和每个Ti in Args应满足MoveConstructible要求。INVOKE (DECAY_COPY(std::forward<F>(f)), DECAY_COPY(std::forward<Args>(args))...)(20.8.2))必须是一个有效的表达式"。

表达式DECAY_COPY(x)在30.2.6中定义:

"在这个子句的几个地方使用了DECAY_COPY(x)操作。所有这些用法都意味着调用函数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类型cl1到类型cl2的转换操作符。为了验证这一点，std::thread的转发机制显然会生成对cl1的右值引用，并试图从中获取c2的实例。这是失败的，因为右值引用不能绑定到转换构造函数中的非const左值引用。

如果你将构造函数的签名从cl1(cl2& ob1)更改为cl1(cl2 const& ob1)，它可以在GCC 4.7.2中工作，因为右值引用可以绑定到左值引用到const。

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