複製構造錯誤，沒有明顯的副本

Question

我試圖從頭開始構建一些簡單的日誌記錄功能，像'< <'這樣的接口，並且遇到了一些編譯器問題。這裏是我的代碼的基礎：

#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"; 
} 

我得到的錯誤是：

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 

從我已經能夠到目前爲止發現的錯誤，因爲你不能在ostringstream上調用複製構造函數。據我所知，我沒有直接調用拷貝構造函數，也沒有拷貝Buffer，只是在'return'語句中構造它。

另一個有趣的事情是，這個代碼在Visual Studio 2010下很好地編譯，在將它集成到我的應用程序（它使用GCC 4.6.3編譯）之前，它將它敲掉。

我是否正確解釋了問題，如果是的話 - 隱式副本在哪裏，以及如何消除它？

Answer 1

第一個問題是std::ostringstream is non-copyable，並且您在創建副本Buffer時嘗試創建副本。

爲什麼要在這個return聲明創建的Buffer副本原因：

return Buffer(*this); 

那是你Buffer類沒有一個隱含產生的移動構造函數。而它沒有的原因是Buffer有一個用戶定義的析構函數，它禁止移動構造函數的隱式生成。

您可以定義一個明確的：

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

但是你應該知道的事實標準庫的一些實施不完全兼容，並沒有實現move constructor of std::ostringstream。