3

我能够为一个简单的结构定义输出流运算符，但是，不能为 std::array 定义。以下代码无法编译。出了什么问题，我该如何解决？

#include <array>
#include <iostream>
#include <boost/log/core.hpp>
#include <boost/log/trivial.hpp>

using hash_t = std::array< unsigned char, 32 >;

std::ostream& operator<< ( std::ostream& os, hash_t const& arr )
{
    os << "ole!";
    return os;
}

int main(int, char*[])
{
    hash_t arr;
    std::cerr << arr << std::endl; // complies cleanly
    BOOST_LOG_TRIVIAL(debug) << arr; // Error
    return 0;
}

错误来了。

GCC（提升 1.55，gcc-4.9.2）：

In file included from /usr/include/boost/log/sources/record_ostream.hpp:31:0,
                 from /usr/include/boost/log/trivial.hpp:23,
                 from trival.cpp:4:
/usr/include/boost/log/utility/formatting_ostream.hpp: In instantiation of ‘boost::log::v2s_mt_posix::basic_formatting_ostream<CharT, TraitsT, AllocatorT>& boost::log::v2s_mt_posix::operator<<(boost::log::v2s_mt_posix::basic_formatting_ostream<CharT, TraitsT, AllocatorT>&, const T&) [with CharT = char; TraitsT = std::char_traits<char>; AllocatorT = std::allocator<char>; T = std::array<unsigned char, 32ul>]’:
trival.cpp:18:30:   required from here
/usr/include/boost/log/utility/formatting_ostream.hpp:710:19: error: cannot bind ‘boost::log::v2s_mt_posix::basic_formatting_ostream<char>::ostream_type {aka std::basic_ostream<char>}’ lvalue to ‘std::basic_ostream<char>&&’
     strm.stream() << value;
                   ^
In file included from /usr/include/c++/4.9/iostream:39:0,
                 from trival.cpp:2:
/usr/include/c++/4.9/ostream:602:5: note: initializing argument 1 of ‘std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = std::array<unsigned char, 32ul>]’
     operator<<(basic_ostream<_CharT, _Traits>&& __os, const _Tp& __x)

Clang 错误（提升 1.64，clang-800.0.42.1）：

In file included from /usr/local/include/boost/log/trivial.hpp:23:
In file included from /usr/local/include/boost/log/sources/record_ostream.hpp:36:
/usr/local/include/boost/log/utility/formatting_ostream.hpp:878:19: error: invalid operands to binary expression ('ostream_type' (aka
      'basic_ostream<char, std::__1::char_traits<char> >') and 'std::__1::array<unsigned char, 32>')
    strm.stream() << value;
    ~~~~~~~~~~~~~ ^  ~~~~~
/usr/local/include/boost/log/sources/record_ostream.hpp:390:51: note: in instantiation of function template specialization
      'boost::log::v2_mt_posix::operator<<<boost::log::v2_mt_posix::basic_formatting_ostream<char, std::__1::char_traits<char>, std::__1::allocator<char> >,
      std::__1::array<unsigned char, 32> >' requested here
    static_cast< formatting_ostream_type& >(strm) << value;
                                                  ^
/Users/adam/GitPen/BoostLog/trival/trival.cpp:18:27: note: in instantiation of function template specialization
      'boost::log::v2_mt_posix::operator<<<boost::log::v2_mt_posix::basic_record_ostream<char>, std::__1::array<unsigned char, 32> >' requested here
        BOOST_LOG_TRIVIAL(debug) << arr;
4

1 回答 1

3

你可能不喜欢这个答案。

要让它在此刻正常工作，您需要将您的添加operator<<到namespace std：

namespace std{
   std::ostream& operator<< ( std::ostream& os, const hash_t& arr )
   {/*...*/}
}

operator<<由于 ADL 的工作方式，它只会考虑namespace std

你不会喜欢它，因为添加这个函数namespace std是不合法的：

[命名空间.std]

std如果 C++ 程序将声明或定义添加到命名空间或命名空间 std 内的命名空间，则 C++ 程序的行为是未定义的，除非另有说明。只有当声明依赖于用户定义的类型并且特化满足原始模板的标准库要求并且没有明确禁止时，程序才能将任何标准库模板的模板特化添加到命名空间 std。

std也许最简单的做法是继承（而不做其他任何事情）类型：

struct hash_t : std::array< unsigned char, 32 >{};
于 2018-01-27T17:03:17.467 回答