Mar*_*s V 6 c++ logging boost boost-log
以下代码与boost 1.57一样正常工作:
#include <iostream>
#include <boost/log/trivial.hpp>
struct Foo
{
int d=1;
};
std::ostream& operator<<(std::ostream& out, const Foo& foo)
{
out << "Foo: " << foo.d;
return out;
}
int main()
{
BOOST_LOG_TRIVIAL(info) << Foo();
return EXIT_SUCCESS;
}
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使用boost 1.59相同的代码失败.第一个gcc错误消息是:
错误:'operator <<'不匹配(操作数类型为'boost :: log :: v2s_mt_posix :: basic_record_ostream'和'Foo')
文档和发行说明均未记录需要更改的内容.
实时版本
看起来问题出在enable_if_formatting_ostream
结构中。它已添加到此提交中。看起来像
template< typename StreamT, typename R >
struct enable_if_formatting_ostream {};
template< typename CharT, typename TraitsT, typename AllocatorT, typename R >
struct enable_if_formatting_ostream< basic_formatting_ostream< CharT, TraitsT, AllocatorT >, R > { typedef R type; };
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现在operator <<
是
template< typename StreamT, typename T >
inline typename boost::log::aux::enable_if_formatting_ostream< StreamT, StreamT& >::type
operator<< (StreamT& strm, T const& value)
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之前是
template< typename CharT, typename TraitsT, typename AllocatorT, typename T >
inline basic_formatting_ostream< CharT, TraitsT, AllocatorT >&
operator<< (basic_formatting_ostream< CharT, TraitsT, AllocatorT >& strm, T const& value)
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因为record_ostream
从编译器派生可以找到重载,但现在不能,因为使用了 SFINAE 并且 struct仅在使用时才formatting_ostream
会有typedef。这可以解决这种情况。type
formatting_ostream