AMDG fmingu wrote:

By using Dev-C++, I used lambda expression, I want to know that if I white: typedef std::vector<int> vect1; std::vector<vect1> vect2; vect2 vect2Instance; then std::for_each(vect2Instance.begin(),vect2Instance.end(), std::for_each(_1.begin(),_1.end(),_1+=2)); is legal or not? And if legal ,Does the first two _1's mean a number in vect2 (i.e. a vect1) and the last _1 is a number in vect1(i.e. a int)? Can any one with kindness help me?

No. It isn't legal. std::for_each cannot return a function object. Boost.Lambda provides lazy wrappers for some algorithms. Here's a complete sample. Note that you can't use boost::lambda::call_begin because it always returns a const_iterator. (Which I will fix shortly). #include <boost/lambda/lambda.hpp> #include <boost/lambda/algorithm.hpp> #include <boost/lambda/bind.hpp> #include <boost/range/iterator.hpp> #include <boost/range/begin.hpp> #include <boost/range/end.hpp> #include <vector> using namespace boost::lambda; struct begin { template<class Args> struct sig { typedef typename boost::range_iterator< typename boost::tuples::element<1, Args>::type

::type type; }; template<class T> typename boost::range_iterator<T>::type operator()(T& t) const { return(boost::begin(t)); } };

struct end { template<class Args> struct sig { typedef typename boost::range_iterator< typename boost::tuples::element<1, Args>::type

::type type; }; template<class T> typename boost::range_iterator<T>::type operator()(T& t) const { return(boost::end(t)); } };

int main() { std::vector<std::vector<int> > vec(4, std::vector<int>(4)); std::for_each(vec.begin(), vec.end(), bind(ll::for_each(), bind(begin(), _1), bind(end(), _1), protect(_1 += 2))); } In Christ, Steven Watanabe