If I have a type, T
, that looks like Foo<mpl::_1>
, I can do mpl::apply<T, int>::type
to get Foo<int>
.
But if T
is a complete type like, say, Foo<int>
, then mpl::apply<T, int>::type
will not compile.
I want to write a metafunction that will apply a type if possible, otherwise return the type. So something like:
template <typename Partial, typename T>
struct maybe_apply
: eval_if_c<??????,
mpl::apply<Partial, T>,
mpl::identity<Partial>>
{ };
What can I put in the ???s so that this does what I want?
Disclaimer: I'm far from an expert on MPL, so I can't promise that this is the best way to solve this (or even if it is correct, it seems to work at least).
According to the documentation, the first parameter/argument to mpl::apply
needs to be a Lambda Expression, and that can be either a Metafunction Class or a Placeholder Expression. A quick google search led me to this post. According to that post, mpl::is_lambda_expression
allows you to determine if a type is a Placeholder Expression. In Boost.TTI (that is in boost since version 1.54) you can find a metafunction that does exactly what you want. This metafunction is boost::tti::detail::is_lambda_expression
and can be found in boost/tti/detail/dlambda.hpp
. In the example below I have used the exact same macro that TTI uses to find if a type is a Metafunction Class.
#include <iostream>
#include <typeinfo>
#include <boost/utility.hpp>
#include <boost/mpl/apply.hpp>
#include <boost/mpl/eval_if.hpp>
#include <boost/mpl/identity.hpp>
#include <boost/mpl/plus.hpp>
namespace mpl=boost::mpl;
/* //This is another way to do it
template <typename T, typename Enable=void>
struct is_apply_able : mpl::false_
{};
template <typename T>
struct is_apply_able<T,typename boost::enable_if<mpl::is_lambda_expression<T> >::type> : mpl::true_
{};
BOOST_MPL_HAS_XXX_TEMPLATE_NAMED_DEF(is_metafunction_class, apply, false)
template <typename T>
struct is_apply_able<T,typename boost::enable_if<is_metafunction_class<T> >::type> : mpl::true_
{};*/
BOOST_MPL_HAS_XXX_TEMPLATE_NAMED_DEF(is_metafunction_class, apply, false)
template <typename T>
struct is_apply_able : mpl::or_<is_metafunction_class<T>,mpl::is_lambda_expression<T> >
{};
struct plus_two
{
template <typename Number>
struct apply
{
typedef typename mpl::plus<Number,mpl::int_<2> >::type type;
};
};
template <typename T>
struct Foo
{};
template <typename Partial, typename T>
struct maybe_apply
: mpl::eval_if<is_apply_able<Partial>,
mpl::apply<Partial, T>,
mpl::identity<Partial> >
{ };
int main()
{
std::cout << typeid(maybe_apply<Foo<mpl::_1>,int>::type).name() << std::endl;
std::cout << typeid(maybe_apply<plus_two,mpl::int_<1> >::type).name() << std::endl;
std::cout << typeid(maybe_apply<Foo<float>,int>::type).name() << std::endl;
}