Re: [Boost-users] [lambda] How to write a lambda functor that returns a new functor

Subject: Re: [Boost-users] [lambda] How to write a lambda functor that returns a new functor
From: Sebastian Theophil (stheophil_at_[hidden])
Date: 2011-03-10 05:12:30

Steven,

> 
> On 03/09/2011 06:11 AM, Sebastian Theophil wrote:
> > How can I write a lambda expression with two placeholders, one for
the
> > callable object, and one for the function argument, such that
> > supplying the callable object first returns a unary function?
> 
> In this case, boost::lambda::_1 by itself does what you want, because
the
> callable object itself is a unary function.

That is true of course, I was trying to solve a slightly more
complicated problem though. I have solved the problem finally, although
I hoped there was a more elegant solution:

struct FCreateBind {
        typedef boost::_bi::bind_t<bool, Callable,
boost::_bi::list2<boost::arg<1>, boost::arg<2> >  > result_type;
        result_type operator()( Callable const& c ) const {
            return boost::bind<bool>(c, _1);
        }
};
BOOST_AUTO(generate, boost::bind(FCreateBind(), _1));

BOOST_AUTO(fn, generate(Callable());
bool b = fn(Arg());

Of course, in this simple example I could just write
BOOST_AUTO(generate, boost::lambda::_1) since Callable itself is the
callable object. But I was looking for a way to set the arguments of
Callable beforehand so the generated function fn is a nullary function.
This solution would let me do this inside FCreateBind.

FCreateBind can probably be eliminated as well, but I have not yet
figured out how to define the pointer to the overloaded global function
boost::bind.

Regards
Sebastian

--
Sebastian Theophil | stheophil_at_[hidden]
Senior Software Engineer
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