没有明显副本的复制构造函数错误

Question

我正在尝试从头开始构建一些简单的日志记录功能，使用类似“

#include <string.h>
#include <sstream>
#include <iostream>

class Logger
{
public:
    class Buffer
    {
    public:
        Buffer(Logger &parent) : mPar(parent)
        {
        }

        ~Buffer(void)
        {
            mPar.endline(os);
        }

        template<class T>
        Buffer& operator<<(const T& x)
        {
            os << x;
            return *this;
        }

        Logger& mPar;
        std::ostringstream os;
    };

    Buffer write(void)
    {
        return Buffer(*this);
    }

    void endline(std::ostringstream& os)
    {
        std::cout << os.str() << std::endl;
    }
};

int main(int argc, char **argv)
{
    Logger log;

    log.write() << "fred" << 3 << "bob";
}

我得到的错误是：

In file included from /usr/include/c++/4.6/ios:45:0,
                 from /usr/include/c++/4.6/istream:40,
                 from /usr/include/c++/4.6/sstream:39,
                 from test.cpp:2:
/usr/include/c++/4.6/bits/ios_base.h: In copy constructor ‘std::basic_ios<char>::basic_ios(const std::basic_ios<char>&)’:
/usr/include/c++/4.6/bits/ios_base.h:788:5: error: ‘std::ios_base::ios_base(const std::ios_base&)’ is private
/usr/include/c++/4.6/bits/basic_ios.h:64:11: error: within this context
In file included from test.cpp:2:0:
/usr/include/c++/4.6/sstream: In copy constructor ‘std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)’:
/usr/include/c++/4.6/sstream:373:11: note: synthesized method ‘std::basic_ios<char>::basic_ios(const std::basic_ios<char>&)’ first required here
/usr/include/c++/4.6/streambuf: In copy constructor ‘std::basic_stringbuf<char>::basic_stringbuf(const std::basic_stringbuf<char>&)’:
/usr/include/c++/4.6/streambuf:782:7: error: ‘std::basic_streambuf<_CharT, _Traits>::basic_streambuf(const __streambuf_type&) [with _CharT = char, _Traits = std::char_traits<char>, std::basic_streambuf<_CharT, _Traits>::__streambuf_type = std::basic_streambuf<char>]’ is private
/usr/include/c++/4.6/sstream:60:11: error: within this context
/usr/include/c++/4.6/sstream: In copy constructor ‘std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)’:
/usr/include/c++/4.6/sstream:373:11: note: synthesized method ‘std::basic_stringbuf<char>::basic_stringbuf(const std::basic_stringbuf<char>&)’ first required here
test.cpp: In copy constructor ‘Logger::Buffer::Buffer(const Logger::Buffer&)’:
test.cpp:8:8: note: synthesized method ‘std::basic_ostringstream<char>::basic_ostringstream(const std::basic_ostringstream<char>&)’ first required here
test.cpp: In member function ‘Logger::Buffer Logger::write()’:
test.cpp:33:22: note: synthesized method ‘Logger::Buffer::Buffer(const Logger::Buffer&)’ first required here

从我目前能够找到的情况来看，错误是因为您无法在 ostringstream 上调用复制构造函数。据我所知，我没有直接调用复制构造函数，也没有复制 Buffer，只是在 'return' 语句中构造它。

另一个有趣的事情是，这段代码在 Visual Studio 2010 下编译得很好，我在将它集成到我的应用程序（使用 GCC 4.6.3 编译）之前将其敲掉。

我是否正确解释了这个问题，如果是，隐式副本在哪里，如何消除它？

Answer 1

第一个问题是std::ostringstream is non-copyable，您在创建Buffer 的副本时尝试创建它的副本。

您在此return 语句中创建Buffer 副本的原因：

return Buffer(*this);

你的类Buffer 没有隐式生成的移动构造函数吗？而它之所以没有，是因为Buffer有一个用户自定义的析构函数，它抑制了move构造函数的隐式生成。

你可以明确定义一个：

    Buffer(Buffer&& b)
        :
        mPar(b.mPar),
        os(std::move(b.os))
    {
    }

但您应该意识到标准库的某些实现并不完全兼容，并且没有实现move constructor of std::ostringstream。