Stringstream错误:无法访问类'std :: basic_ios <_Elem,_Traits>'中声明的私有成员

Question

在创建一个简单的异常类扩展(我可以更容易地构造错误消息)时,我将错误分解为以下简单代码:

#include <sstream>
#include <string>
class myCout {
public:
    std::stringstream ssOut;    // Removing this gets rid of error
    template <typename T> myCout& operator << (const T &x) {
        // Do some formatting
        return *this;
    }
};

class myErr : public myCout {
public:
    using myCout::operator<<;
};

int main(int argc, const char* argv[]) {
    throw myErr() << "ErrorMsg" << 1;
    myCout() << "Message Will Be Formatted";
    return 0;
}

在编译时,会产生以下错误:

1>C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\include\sstream(724): error C2248: 'std::basic_ios<_Elem,_Traits>::basic_ios' : cannot access private member declared in class 'std::basic_ios<_Elem,_Traits>'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\include\ios(176) : see declaration of 'std::basic_ios<_Elem,_Traits>::basic_ios'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>
1>          ]
1>          This diagnostic occurred in the compiler generated function 'std::basic_stringstream<_Elem,_Traits,_Alloc>::basic_stringstream(const std::basic_stringstream<_Elem,_Traits,_Alloc> &)'
1>          with
1>          [
1>              _Elem=char,
1>              _Traits=std::char_traits<char>,
1>              _Alloc=std::allocator<char>
1>          ]

(实际上它更复杂并且扩展了类似的东西std::runtime_error)

我已经看到以前的答案表明问题产生于不通过引用传递流,但我看不出我不是.

评论std::stringstream ssOut解决问题的方法.为什么,以及如何解决潜在问题？

Answer 1

你是按值抛出异常,这确实是推荐的做法.但是,这意味着异常将作为throw语句的一部分进行复制,因此它必须具有可访问的复制构造函数.因为你有一个不可复制的成员(std::stringstream),你必须提供自己的副本ctor.