Re: [Boost-users] nest mpl::lambda

Subject: Re: [Boost-users] nest mpl::lambda
From: Jeffrey Lee Hellrung, Jr. (jeffrey.hellrung_at_[hidden])
Date: 2013-02-23 18:56:57

On Sat, Feb 23, 2013 at 2:32 AM, Niitsuma Hirotaka <
hirotaka.niitsuma_at_[hidden]> wrote:

> I would like to use mpl::lambda in nest.
>
> In this case works fine.
> ---------
>   typedef boost::mpl::lambda<boost::mpl::max<boost::mpl::_1,boost::mpl::_2>
> > my_max;
>   typedef boost::mpl::fold<
>       boost::mpl::vector_c<int,1,7,0,-2,5,-1>
>     , boost::mpl::int_<6>
>     //, boost::mpl::max<boost::mpl::_1,boost::mpl::_2>
>     , my_max
>     >::type r2;
>
>   BOOST_MPL_ASSERT(( boost::is_same< r2, boost::mpl::integral_c<int, 7> >
> ));
> ---------
>
> However
>
> ---------
>
>
>   typedef boost::mpl::lambda<
>     boost::mpl::fold<
>       boost::mpl::_1
>       ,boost::mpl::front<boost::mpl::_1>
>       ,my_max
>       >::type >::type my_max_types;
>
>   typedef my_max_types::apply<boost::mpl::vector_c<int,1,7,0,-2,5,-1>
> >::type r3;
>
>
>   BOOST_MPL_ASSERT(( boost::is_same< r3, boost::mpl::integral_c<int, 7> >
> ));
>   ///error => becomes boost::mpl::integral_c<int, 1>
> ---------
>
>
> How to define such nest lambda in mpl?
>

Uh, I see a red flag in that you don't grab the nested type typedef from
your mpl::lambda invocations, i.e., I think you should have

typedef mpl::lambda< mpl::max< mpl::_1, mpl::_2 > >::type my_max;

Incidentally, you could also do

typedef mpl::quote2< mpl::max > my_max

(Well, assuming mpl::max doesn't have some hidden default parameters...)

Without grabbing the type typedef in your mpl::lambda invocation, your use
of mpl::_1 and mpl::_2 are "exposed" to premature substitution.

- Jeff