c++ - boost::iostreams::copy() 关闭源而不是接收器

Question

我正在尝试使用 boost::iostreams 来压缩数据。

copy() 的文档表示它的两个参数最后通过调用模板函数关闭 close()在他们两个上。我的测试代码是:

#include <iostream>
#include <fstream>

#include <boost/iostreams/filtering_streambuf.hpp>
#include <boost/iostreams/copy.hpp>
#include <boost/iostreams/filter/gzip.hpp>

using namespace std;

int main(void)
{
    ifstream ifs("output", ios::binary);
    ofstream ofs("output.boost.gz", ios::binary);

    boost::iostreams::filtering_streambuf<boost::iostreams::output> out;

    out.push(boost::iostreams::gzip_compressor());
    out.push(ofs);

    cout << (ifs.is_open() ? "ifs is opened" : "ifs not opened") << endl;
    cout << (ofs.is_open() ? "ofs is opened" : "ofs not opened") << endl;

    boost::iostreams::copy(ifs, out);

    cout << (ifs.is_open() ? "ifs is opened" : "ifs not opened") << endl;
    cout << (ofs.is_open() ? "ofs is opened" : "ofs not opened") << endl;

    return 0;
}

这个测试输出:

ifs is opened
ofs is opened
ifs not opened
ofs is opened

可以看到ofs还是打开的。我的问题是:为什么？当传递一个 filtering_streambuf 对象时，boost::iostreams::close() 做了什么？

Answer 1

有趣。

走下兔子洞^[1] 结果是close_impl<any_tag>最终到达 filtering_streambuf 内的 chain_buf 深处的 ofstream。实现如下:

template<>
struct close_impl<any_tag> {
    template<typename T>
    static void close(T& t, BOOST_IOS::openmode which)
    {
        if (which == BOOST_IOS::out)
            iostreams::flush(t);
    }

    template<typename T, typename Sink>
    static void close(T& t, Sink& snk, BOOST_IOS::openmode which)
    {
        if (which == BOOST_IOS::out) {
            non_blocking_adapter<Sink> nb(snk);
            iostreams::flush(t, nb);
        }
    }
};

因此，如您所见，记录的行为实际上只是刷新了链接的输出流缓冲区(在该调用之前，包含实体上还有一个同步，IIRC) .

我完全同意，这本可以更加明确。

阅读决定特化的 TMP 代码:

template<typename T>
struct close_tag {
    typedef typename category_of<T>::type             category;
    typedef typename detail::unwrapped_type<T>::type  unwrapped;
    typedef typename
            iostreams::select<
                mpl::not_< is_convertible<category, closable_tag> >,
                any_tag,
                mpl::or_<
                    is_boost_stream<unwrapped>,
                    is_boost_stream_buffer<unwrapped>
                >,
                close_boost_stream,
                mpl::or_<
                    is_filtering_stream<unwrapped>,
                    is_filtering_streambuf<unwrapped>
                >,
                close_filtering_stream,
                mpl::or_<
                    is_convertible<category, two_sequence>,
                    is_convertible<category, dual_use>
                >,
                two_sequence,
                else_,
                closable_tag
            >::type type;
};

我想到了几种解决方法:

1. 定义 close_tag<> 的特化对于 std::ofstream实际上返回一个不同的标签并使其关闭(我建议不要这样做，因为它可能会因违背 Boost Iostreams 开发人员所持有的假设而产生意想不到的效果)

2. 为输出流使用 boost 类:

#include <iostream>
#include <fstream>

#include <boost/iostreams/filtering_streambuf.hpp>
#include <boost/iostreams/copy.hpp>
#include <boost/iostreams/device/file.hpp>
#include <boost/iostreams/filter/gzip.hpp>

using namespace std;

int main(void)
{
    cout << boolalpha;

    ifstream ifs("output", ios::binary);
    boost::iostreams::file_sink ofile("output.boost.gz");

    boost::iostreams::filtering_streambuf<boost::iostreams::output> out;
    out.set_auto_close(true);

    out.push(boost::iostreams::gzip_compressor());
    out.push(ofile);

    cout << "out.is_complete(): " << out.is_complete() << endl;
    cout << "ifs.is_open()? "     << ifs.is_open()     << endl;
    cout << "ofile.is_open()? "   << ofile.is_open()   << endl;

    boost::iostreams::copy(ifs, out);

    cout << "out.is_complete(): " << out.is_complete() << endl;
    cout << "ifs.is_open()? "     << ifs.is_open()     << endl;
    cout << "ofile.is_open()? "   << ofile.is_open()   << endl;
}

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^[1] 这是一个大得惊人的兔子洞，我必须补充一下。我想知道所有这些通用性到底有什么好处