Hello,
for formatting output I wrote the following classes:
---8<---
namespace /* anonymous */ {
namespace mpl = ::boost::mpl;
using ::boost::enable_if;
using ::boost::xpressive::sregex;
using ::boost::xpressive::as_xpr;
template<typename StringT>
struct is_string : mpl::false_ { };
template<>
struct is_stringstd::string : mpl::true_ { };
template
struct formatter
{
formatter( const Tp& data )
: output( data )
{ }
std::ostream& write_on( std::ostream& os ) const
{
os << output;
return os;
}
private:
const Tp& output;
};
template <typename Tp>
struct formatter::type>
{
formatter( const Tp& data )
: output( data ),
re( as_xpr("\r") ),
format("\n")
{ }
std::ostream& write_on( std::ostream& os ) const
{
using ::boost::xpressive::regex_replace;
std::ostream_iteratorstd::ostream::char_type out_iter( os );
regex_replace( out_iter, output.begin(), output.end(), re,
format );
return os;
}
private:
const Tp& output;
const sregex re;
const std::string format;
};
}
--->8---
Mainly, this replaces all '\r' by '\n' in all string in the given data,
e.g.:
---8<---
template<typename DataT>
void foo::write_key_data_pair( std::ostream& os,
const std::string& key, const DataT& data ) const
{
os << "[" << key << "]"
<< " "
;
formatter<DataT>( data ).write_on( os );
os << std::endl
;
}
--->8---
Anyway, overloading global operator<<() as
template<typename T>
static inline
std::ostream& operator<<( std::ostream& os, const T& rhs )
{
return rhs.write_on( os );
}
would force all to have a write_on member function, isn't it. Is there a
way around and even use operator<<() notation ??
Thanks,
Olaf
