2009/7/21 fmingu <fmingu@163.com>
Oh.Please do not play joke on me. I am doing my work seriously. I really want the answer on lambda expression in order to know lambda expression further. Can any one with kindness help me?
I am using Dev-C++ 4.9.8.0,with lambda expression:
..................... class A{ ............. int result; double function2(const int& n) ......................... double function3(const int& Q,const int& N) ...................... double function1(const int& q, int& n){ double result=0.0; int loopvalue=0, totaltime=n; n%=q; ++thenumbern; ( for_loop(var(loopvalue)=0,var(loopvalue)<=var(totaltime),++var(loopvalue), var(result)+=( bind(&A::function2,_2)*bind(&A::function3,_1,_2))))(q,n); return result; } But the complier told me that: ............................... \boost\lambda\detail\select_functions.hpp instantiated from `Arg::sig<boost::tuples::tuple<A&,................. instantiated from `RET boost::lambda::lambda_functor_base<boost::lambda::forloop_action, Args>::call(A&, B&, C.................... \boost\lambda\detail\lambda_functors.hpp instantiated from `T::sig<boost::tuples::tuple<A&, B&, boost::tuples::null_type, boost::tuples::null_type, ............... instantiated from here
By the way, is the result in var(result) the double result in function1 or the int result in class A? Can any one with kindness help me?
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You've forgotten that member functions need an object reference...change bind(&A::function2,_2) to bind(&A::function2,this,_2) and bind(&A::function3,_1,_2) to bind(&A::function3,this,_1,_2) HOWEVER - as Roman was alluding - just because you CAN do something (like this loop) in lambda doesn't mean it's a good idea to... Stuart Dootson