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

9 Mar 2011


      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
struct Arg {
};
struct Callable : std::unary_function  {
     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::value));
     Arg a;
     bool b = fn(a);
     _ASSERT(b==true);
}
In Christ,
Steven Watanabe