Hi, 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 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))); 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); } Thanks, Sebastian -- Sebastian Theophil | stheophil@think-cell.com Senior Software Engineer think-cell Software GmbH | Chausseestr. 8/E | 10115 Berlin | Germany http://www.think-cell.com | phone +49 30 666473-10 | US phone +1 800 891 8091 Amtsgericht Berlin-Charlottenburg, HRB 85229 | European Union VAT Id DE813474306 Directors: Dr. Markus Hannebauer, Dr. Arno Schoedl