为什么不能构造包含ostringstream成员的对象？

Question

我有以下类示例,从较大的项目中简化.它基于一个日志框架,它使用记录器的作用域来终止析构函数中的日志条目.

下面的代码将无法编译,因为构造函数是一个隐式删除的函数(编辑:不是真),这似乎与std::ostringstream对象有关.我对此感到困惑,因为我认为我应该能够直接构造一个std::ostringstream,这意味着我应该能够直接构造一个Container对象.

#include <iostream>
#include <sstream>

class Container {
  public:
    std::ostringstream  bufferStream;

  public:
    Container();    // constructor
    ~Container();
};

Container::Container() {
    bufferStream << "Hello ";
}

Container::~Container() {
    std::cout << bufferStream.str() << " [end]" << std::endl;
}

// === Main method ===

int main() {

    Container().bufferStream << "world";   // works fine

    {                                      // causes tons of compiler errors
        Container cont = Container();
        cont.bufferStream << "world!";
    }

    return 0;
}

请注意,标记为"正常工作"的行就是这样做的.它似乎实例化了一个匿名Container对象,它包含一个新的std::ostringstream,可以直接访问输出"world".它Container自己创建消息的"Hello"部分,它的析构函数刷新缓冲区.

为什么Container命名和保存对象的第二部分不能正确运行？以下是我得到的错误示例:

error.cpp: In function ‘int main()’:
error.cpp:28:36: error: use of deleted function ‘Container::Container(const Container&)’
         Container cont = Container();
                                    ^
error.cpp:4:7: note: ‘Container::Container(const Container&)’ is implicitly deleted because the default definition would be ill-formed:
 class Container {
       ^
error.cpp:4:7: error: use of deleted function ‘std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)’
In file included from error.cpp:2:0:
/usr/include/c++/4.8/sstream:387:11: note: ‘std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)’ is implicitly deleted because the default definition would be ill-formed:
     class basic_ostringstream : public basic_ostream<_CharT, _Traits>

... 等等.

Answer 1

这样可以正常工作:

Container cont;
cont.bufferStream << "world!";

但是这个:

Container cont = Container();

涉及复制构造函数.std::ostringstream不是可复制构造的,这使得Container不可复制构造,因此错误消息谈论Container::Container(const Container&)由于std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)被隐式删除而如何被隐式删除.

请注意,即使此副本将被省略,复制/移动省略的要求是必须可以开始复制/移动.