我的目标是要有几个流。您可以在下面查看工作代码。

我正在尝试使用可变参数模板,但失败了。下一个代码如何“变”?

有几个非常相似的变量,因此我想可以使用可变参数模板将其重写,但我不知道如何。

template<typename T>
struct IsOn
{
 T *pt;
 bool isOn;
 IsOn(T& t, bool b):pt(&t),isOn(b) {}
};

struct tmy
{
 vector<IsOn<ostream>> v0;
 vector<IsOn<ofstream>> v1;
 vector<IsOn<stringstream>> v2;
};

template<typename T>
tmy& operator<<(tmy& rt,T& t) {
 int len;
 len=rt.v0.size();
 for(int i=0; i<len;++i) if(rt.v0[i].isOn) (*rt.v0[i].pt)<<t;
 len=rt.v1.size();
 for(int i=0; i<len;++i) if(rt.v1[i].isOn) (*rt.v1[i].pt)<<t;
 len=rt.v2.size();
 for(int i=0; i<len;++i) if(rt.v2[i].isOn) (*rt.v2[i].pt)<<t;
 return rt;
}


int main(int argc, char** argv) {
 tmy my;
 my.v0.push_back(IsOn<ostream>(cout, true));
 my.v0.push_back(IsOn<ostream>(cerr, false));
 my.v0.push_back(IsOn<ostream>(clog, true));
 my<<"hi twice!";
}

感谢您的尝试!

ps我知道这里存在boost::tee,但是我有稍微不同的问题,可以在这里阅读:How to declare an "implicit conversion" in a variadic template?

最佳答案

好的,我有两件事可以解决,但我认为这没有什么意义:

#include <iostream>

template<typename ... Streams>
struct StreamCont
{
};

template<typename Stream>
struct StreamCont<Stream>
{
    Stream & stream;
    StreamCont(Stream & st) :  stream(st) {};
};

template<typename Stream, typename ... Next>
struct StreamCont<Stream, Next...>
{
    Stream & stream;
    StreamCont<Next...> next;

    StreamCont(Stream & st, Next&... next) :  stream(st), next(next...) {};
};

template<typename Stream, typename Arg>
StreamCont<Stream>& operator<<(StreamCont<Stream> & str, Arg arg)
{
    str.stream << arg;
    return str;
};

template<typename ... Streams, typename Arg>
StreamCont<Streams...>& operator<<(StreamCont<Streams...> & str, Arg arg)
{
    str.stream << arg;
    str.next << arg;
    return str;
};

/* std::endl signature:
template< class CharT, class Traits >
std::basic_ostream<CharT, Traits>& endl( std::basic_ostream<CharT, Traits>& os );

so this only works if all streams are equal
*/

template<typename Stream>
StreamCont<Stream>& operator<<(StreamCont<Stream> & str, Stream&(*func)(Stream&) )
{
    func(str.stream);
    return str;
};


template<typename First, typename ... Streams>
StreamCont<First, Streams...>& operator<<(StreamCont<First, Streams...> & str, First&(*func)(First&) )
{
    func(str.stream);
    str.next << func;
    return str;
};


int main()
{
    StreamCont<std::ostream, std::ostream, std::ostream>
        multi_stream(std::cout, std::cerr, std::clog);

    multi_stream << 42 << std::endl;
    return 0;
}

对数组执行相同的操作可能更有意义,即:
#include <iostream>
#include <array>


template<typename Stream, size_t Size, typename Arg>
std::array<Stream*, Size>& operator<<(std::array<Stream*, Size>& str, const Arg &arg)
{
    for (auto  s : str)
        *s << arg;
    return str;
};

template<typename Stream, size_t Size>
std::array<Stream*, Size>& operator<<(std::array<Stream*, Size>& str, Stream& (*func)(Stream&))
{
    for (auto  s : str)
        *s << func;
    return str;
};



int main()
{
    std::array<std::ostream*, 3> strs =  {&std::cout, &std::cerr, &std::clog};

    strs << 42 << std::endl;
    return 0;
}

希望能有所帮助,我不知道您是否可以编写一个自定义的ostream重载(请参见boost.iostreams),而仅使用stream_bufs并将它们组合到您的自定义类中是否更好。

10-02 03:04