具有可变通用引用和复制构造函数的c ++ 11构造函数 [英] c++11 constructor with variadic universal references and copy constructor
问题描述
如果我们也有带有通用引用参数的构造函数,那么如何声明副本构造函数?
How to declare copy constructor, if we have constructor with universal reference arguments, also?
http://coliru.stacked-crooked.com/a/4e0355d60297db57
struct Record{
template<class ...Refs>
explicit Record(Refs&&... refs){
cout << "param ctr" << endl;
}
Record(const Record& other){ // never called
cout << "copy ctr" << endl;
}
Record(Record&& other){ // never called
cout << "move ctr" << endl;
}
};
int main() {
Record rec("Hello");
Record rec2(rec); // do "param ctr"
return 0;
}
根据 std :: tuple的此构造函数列表
http://en.cppreference.com/w / cpp / utility / tuple / tuple [看案例3和8]这个问题在标准库中以某种方式解决了...但是我无法通过stl的代码来解决。
According to this constructor list of std::tuple
http://en.cppreference.com/w/cpp/utility/tuple/tuple [look case 3 and 8] this problem somehow solved in standard library... But I can't get through stl's code.
PS问题与构造函数中的C ++通用引用和返回值优化(rvo)
PPS现在,我只是为真正的EXPLICIT调用添加了其他第一个参数 Record(call_constructor,Refs& ... refs)
。而且我可以手动检测是否只有一个参数,是否为 Record
,然后重定向调用以复制ctr / param ctr,但是....相信没有标准的方法...
P.P.S. For now, I just added additional first param Record(call_constructor, Refs&&... refs)
for really EXPLICIT call. And I can manually detect if we have only one param and if it is Record
, and than redirect call to copy ctr/param ctr, but.... I can't believe there is no standard way for this...
推荐答案
在您的示例中,转发引用与<$ c $一起使用c> Record& 。
In your example, the forwarding reference is used with Record&
.
因此,您可以为 Record&
添加额外的重载(转发给复制构造函数):
So you may add an extra overload for Record&
(to forward to copy constructor):
Record(Record& other) : Record(static_cast<const Record&>(other)) {}
或在具有转发参考的那一个上使用sfinae。
or use sfinae on the one with forwarding reference.
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