Re: [Boost-users] [lamba] Comparing return value of memberfunctionwith ->*

Sorry, I answered your message a little too quick. Actually sample::odd is a member function, so you have to write if_then(bind(&sample::odd, _1) == false, std::cout << _1) &_1->*&sample::odd would work if sample::odd was an attribute, not a member function. Remember to use lambda's bind (i.e. boost::lambda::bind), not boost::bind. Let me know if this works. Regards, rodolfo "Boris" escreveu na mensagem news:op.to4s8qi39dsao3@burk...

On Tue, 13 Mar 2007 00:21:36 +0200, Rodolfo Lima wrote:

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// Why doesn't this compile ... std::for_each(from.begin(), from.end(), if_then((_1 ->* &sample::odd) == false, std::cout << _1));

IMO it should be &1->*&sample::odd

The ->* operator expects a pointer in its left side. Since & has higher priority than ->*, it suffices to add & to _l and it'll work.

I played around some more and even removed the comparison with false. But neither this ...

std::for_each(from.begin(), from.end(), if_then(boost::lambda::_1 ->* &sample::odd, std::cout << boost::lambda::_1));

... nor that ...

std::for_each(from.begin(), from.end(), if_then(&boost::lambda::_1 ->* &sample::odd, std::cout << boost::lambda::_1));

compiles. VC8++ complains about error C2451: conditional expression of type 'boost::lambda::detail::member_pointer_caller' is illegal. Everything works though with boost::lambda::bind. Why I don't know. From the documentation ->* looks like a more readable shortcut than boost::lambda::bind. There seem to be some limitations though (or does it depend on the compiler)?

Boris