如何使用boost :: iostreams管道進入std :: cout

Question

我是boost::iostreams新手，所以這可能是微不足道的：

假設namespace io = boost::iostreams;

這很有效

io::filtering_ostream out(std::cout);
out << "some\nstring\n";

這很有效

std::string result;
io::filtering_ostream out(io::counter() | io::back_inserter(result));
out << "some\nstring\n";

但這不編譯

io::filtering_ostream out(io::counter() | std::cout);
out << "some\nstring\n";

你如何管道進入std::cout ？

Answer 1

用boost :: ref包裝std::cout對我std::cout ：

io::filtering_ostream out(DummyOutputFilter() | boost::ref(std::cout));

有關詳細信息，請參閱pipable 文檔中的note_1 。

Answer 2

為了完整起見，一個簡單的“Sink wrapper”看起來像這樣：

#include <boost/iostreams/concepts.hpp>
#include <boost/iostreams/pipeline.hpp>

template<typename Sink>
class sink_wrapper
    : public boost::iostreams::device<boost::iostreams::output, typename Sink::char_type> {
public:
    sink_wrapper(Sink & sink) : sink_(sink) {}

    std::streamsize write(const char_type * s, std::streamsize n) {
        sink_.write(s, n);
        return n;
    }

private:
    sink_wrapper & operator=(const sink_wrapper &);
    Sink & sink_;
};
BOOST_IOSTREAMS_PIPABLE(sink_wrapper, 1)

template<typename S> sink_wrapper<S> wrap_sink(S & s) { return sink_wrapper<S>(s); }

並且可以像這樣使用：

boost::iostreams::filtering_ostream  out(filter | wrap_sink(std::cout));

Answer 3

那不是你傳遞流的方式。 你必須使用push ：

out.push(std::cout);