Re: [boost] [TypeSort] Automatic type sorting

On 20/03/2015 17:49, costis glynos wrote:

...

You also need to use Boost.Fusion to make it into a tuple at the end, since this is what you want.

Here is an example:

#include #include #include #include #include #include <iostream>

int main() { namespace mpl = boost::mpl; namespace fusion = boost::fusion;

typedef mpl::vector types; typedef mpl::sortmpl::_1 , mpl::sizeof_mpl::_2 > >::type sorted_types; typedef fusion::result_of::as_vector::type tuple;

tuple t; std::cout << sizeof(t) << std::endl; // 8 instead of 12 }

This appears to work.

That's fantastic! Thank you for your recommendation.

You probably need to be able to access the final tuple with the original indices though, which is a bit more involved. #include #include #include #include #include #include #include #include #include #include #include #include #include #include <iostream> int main() { namespace mpl = boost::mpl; namespace fusion = boost::fusion; typedef mpl::vector types; typedef mpl::copy< mpl::zip_view< mpl::vector< types , mpl::range_c > > , mpl::back_inserter< mpl::vector0<> >

::type indexes_types; typedef mpl::sort< indexes_types , mpl::less< mpl::sizeof_< mpl::frontmpl::_1 > , mpl::sizeof_< mpl::frontmpl::_2 > > ::type indexes_types_sorted; typedef mpl::transform< indexes_types_sorted , mpl::frontmpl::_1 ::type sorted_types; typedef mpl::transform< indexes_types_sorted , mpl::backmpl::_1 ::type sorted_indexes; typedef fusion::result_of::as_vector::type tuple;

tuple t; std::cout << sizeof( fusion::at_c::type::value>(t) ) << std::endl; std::cout << sizeof( fusion::at_c::type::value>(t) ) << std::endl; std::cout << sizeof( fusion::at_c::type::value>(t) ) << std::endl; } This could make an interesting addition to Boost.Fusion.