Re: [Boost-users] lambda syntax

From: Aleksey Gurtovoy (agurtovoy_at_[hidden])
Date: 2006-03-01 10:56:07

Lucio Flores writes:
> So as an exercise, I'm trying to use a lambda expression that
> applies the first argument twice on it's second argument.
>
> typedef mpl::apply<_1, mpl::apply<_1, _2>::type>::type twice_type;

I'm afraid this doesn't do what you think. Following a metafunction
with '::type' means invoking it, so the above is basically equivalent
to

    typedef mpl::apply<_1, _2>::type t; 
        // == _1::apply<_2>::type == _2
    
    typedef mpl::apply<_1,t>::type twice_type; 
        // == _1::apply<t>::type == _1::apply<_2>::type
        // == _2

>
> Then I instance this lambda function on boost::add_pointer and int
>
> typedef mpl::apply<twice_type, boost::add_pointer<_1>, int>::type result_type;

.. and, in its turn, this is equivalent to

   typedef mpl::apply<_2, boost::add_pointer<_1>, int>::type result_type;

which gives you the result you refer to below.

>
> But using boost::is_same, I've found that result_type==int, not
> int** as expected. Can someone see why?

HTH,
-- 
Aleksey Gurtovoy
MetaCommunications Engineering