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: Steven Watanabe (watanabesj_at_[hidden])
Date: 2011-03-09 18:28:40

AMDG

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.

> In the example below, generate should be a lambda expression with the
> first placeholder for the callable object itself, and the second
> placeholder for the argument. Calling generate(c) should return a unary
> function that is only missing the function call argument. In fact, it
> somehow returns type bool already, as proved by the static assert.
>
> #include<boost/lambda/bind.hpp>
>
> struct Arg {
> };
>
> struct Callable : std::unary_function<Arg, bool>  {
>      bool operator()( Arg const&  a ) const { return true; }
> };
>
> int main( int argc, const char* argv[] ) {
>      BOOST_AUTO(generate, boost::lambda::bind(boost::lambda::_1,
> boost::lambda::protect(boost::lambda::_1)));
>

generate(c) will end up calling c(boost::lambda::_1).

>      Callable c;
>      BOOST_AUTO(fn, generate(c));
>
>      BOOST_STATIC_ASSERT((boost::is_same<BOOST_TYPEOF(fn),
> bool>::value));
>      Arg a;
>      bool b = fn(a);
>      _ASSERT(b==true);
> }

In Christ,
Steven Watanabe