September 2005 - Boost-users - lists.preview.boost.org

why not boost::bind::_1 ?
by ClassyK 22 Sep '05

22 Sep '05

when include "boost\bind.hpp", why _1 is in global space rather than boost::bind, while in lamba, there is boost::lamba::_1

1 0

Re: [Boost-users] [BGL] detect cycles in a directed graph
by Vladimir Prus 22 Sep '05

22 Sep '05

On Thursday 22 September 2005 10:50, Giulio Veronesi wrote: [replying to list, no need to keep this off-list] > Vladimir Prus ha scritto: > > Giulio Veronesi wrote: > >>assistance in the following problem: "get all the edges of a cycle". > > > > Of which cycle? > > > > If you want to get at least one cycle in a graph that has one, then you > > need to create a visitor that implements "back_edge" method and use > > predecessor map to traverse the cycle in reverse order. > > Yes! Only one cycle. Please, could you give me a pratical example (just > the pseudo code)? I've very little experience with BGL. Do you intend > something like this? > > template <class Edge, class Graph> > void back_edge(Edge e, Graph& g) { > clinks.push_back(TLink(source(e, g), target(e, g))); > } > > where: > > typedef std::pair<int,int> TLink; > vector<TLink> clinks; Something more like: struct vis { vis(vector_property_map<int> predecessor) : m_predecessor(predecessor) {} template<......> void back_edge(Edge e, Graph& g) { vertex_descriptor t = target(e, g); vertex_description c = source(e, g); vector<vertex_descriptor> cycle; cycle.push_back(c); do { c = m_predecessor[c]; cycle.push_back(c); } while(c != t); } }; On construction, you need to pass the same vector_property_map both to the visitor and the depth_first_search algorithm. Yes, this is untested code, sorry :-( - Volodya

4 3

[bind] use with shared_ptr
by Neal Coombes 22 Sep '05

22 Sep '05

Bind is very convenient for use with shared_ptr: struct C { void func() {} }; shared_ptr<C> c(new C); bind(&C::func, c)(); // calls c->func() However I've found a snag when trying to do function composition: class A { shared_ptr<C> func() { return shared_ptr<C>(new C); } }; A a; bind(&C::func, bind(&A::func, a)); // doesn't work, instead must: bind(&C::func, bind(&get_pointer<B>, bind(&A::func, a))); The fact that I can work around this is obviously a good thing. But the work around is significantly less clear, especially when attempting to convince a development team that boost and bind are Good Things(tm) and they should be adopted. Is it horribly difficult to make the simpler composition work (i.e. was it tried and impossible or simply overlooked)? Thanks, Neal

2 2

Re: [Boost-users] A question about boost::lambda::if_ behavior
by Bjorn.Karlsson＠readsoft.com 22 Sep '05

22 Sep '05

From: tom gee [mailto:rockonedge@gmail.com] > Can you please further explain the usage of var and constant . > in the expression if_(_1 < 24.000001 && _1 > 23.999999)[ > cout <<"Bingo:<"<<_1<<">"]) > I understand "Bingo:<" is evaluated immediately, why not ">", > which is not a lambda expression either, is not evaluated along. Actually it is a lambda expression! The reason is the preceding streaming of _1, which implicitly marks the expression as a lambda expression. So, std::cout << "Bingo:<" is not a lambda expression, and is thus evaluated immediately. The last part of the expression is a lambda expression, _1 << ">" (or rather std::cout << _1 << ">"), but it won't be invoked. See http://tinyurl.com/7jq3v for details on how to use constant and var. > Forgive me if this question seems stupid, but I am new to Lambda, > fascinated by its power, but convoluted by its mechanism yet. The question isn't stupid, and you're more than welcome to keep asking. And I agree that Boost.Lambda's power is fascinating. Bjorn Karlsson

1 0

Re: [Boost-users] A question about boost::lambda::if_ behavior
by tom gee 22 Sep '05

22 Sep '05

Hello Bjorn, Thanks for such a quick help. I works now. Can you please further explain the usage of var and constant . in the expression if_(_1 < 24.000001 && _1 > 23.999999)[ cout <<"Bingo:<"<<_1<<">"]) I understand "Bingo:<" is evaluated immediately, why not ">", which is not a lambda expression either, is not evaluated along. Forgive me if this question seems stupid, but I am new to Lambda, fascinated by its power, but convoluted by its mechanism yet. Thanks. -----Original Message----- From: tom gee [mailto:rockonedge@gmail.com] Subject: [Boost-users] A question about boost::lambda::if_ behavior > Can anyone please explain this code: [snip] The problem is that the expression std::cout << constant("Bingo:<") is evaluated immediately. Why? Because it's not a lambda expression. You have two options that will help you turn your if-branch into a lambda expression; either use boost::lambda::var or boost::lambda::constant. Using boost::lambda::var: std::for_each(setFinalVal.begin(),setFinalVal.end(), (std::cout << _1 << " ", if_(_1 < 24.000001 && _1 > 23.999999) [var(std::cout) << "Bingo:<" << _1 << ">"])); Using boost::lambda::constant: std::for_each(setFinalVal.begin(),setFinalVal.end(), (std::cout << _1 << " ", if_(_1 < 24.000001 && _1 > 23.999999) [std::cout << constant("Bingo:<") << _1 << ">"])); Cheers, Bjorn Karlsson

2 1

Re: [Boost-users] algorithms and member functions
by Foster Gareth 22 Sep '05

22 Sep '05

> What's wrong with: > std::for_each > ( as.begin > , as.end() > , bool some_function_keyword(A& a) > { return a.wee() & SOME_FLAG; > } > ); > I think the previous poster (whos comments where appreciated) implied that while some languages (he mentioned Lisp and Perl I think, and I mentioned Java) could/do support this, C++ would struggle to. It would make things hairy for the compiler because a function definition is not preceeded by any keyword. I think you can also do things a bit like this in java ... Interesting perhaps that there is a keyword involved in this case, would this still bog down compilers so much as to outweight the advantages of the sytax style (assuming there are any, I'm happy to be argued out of thiking this kind of coding is neat). button_click_event.set_handler ( new class : public button_click_event_handler { void on_click() { std::cout << "anon_class::on_click() : printing wee" << std::endl; } } ); > I don't know what a good function keyword name would be. > Obviously: fn, function, lambda are in common use everywhere > (and fn is too short for a C++ keyword! :D) > Reuse 'inline'? I suppose inline would make sense, but it might be a bit of a hack to have it used in two senses. I wonder if you can do this stuff in Delphi, they have a function keyword. Thesaurus :) " function noun 1. The proper activity of a person or thing: job, purpose, role, task. See do/not do. 2. A large or important social gathering: affair, celebration, festivity, fete, gala, occasion, party, soiree. Informal do. Slang bash. See group, work/play. verb 1. To react in a specified way: act, behave, operate, perform, work. See action/inaction. 2. To perform a function effectively: go, operate, run, take, work. See thrive/fail/exist. 3. To perform the duties of another: act, officiate, serve. See do/not do, substitute. " Cheers for the responses anyway, Gaz

2 1

Re: [Boost-users] A question about boost::lambda::if_ behavior
by Bjorn.Karlsson＠readsoft.com 22 Sep '05

22 Sep '05

Hello Tom, -----Original Message----- From: tom gee [mailto:rockonedge@gmail.com] Subject: [Boost-users] A question about boost::lambda::if_ behavior > Can anyone please explain this code: [snip] The problem is that the expression std::cout << constant("Bingo:<") is evaluated immediately. Why? Because it's not a lambda expression. You have two options that will help you turn your if-branch into a lambda expression; either use boost::lambda::var or boost::lambda::constant. Using boost::lambda::var: std::for_each(setFinalVal.begin(),setFinalVal.end(), (std::cout << _1 << " ", if_(_1 < 24.000001 && _1 > 23.999999) [var(std::cout) << "Bingo:<" << _1 << ">"])); Using boost::lambda::constant: std::for_each(setFinalVal.begin(),setFinalVal.end(), (std::cout << _1 << " ", if_(_1 < 24.000001 && _1 > 23.999999) [std::cout << constant("Bingo:<") << _1 << ">"])); Cheers, Bjorn Karlsson

2 1

Boost function
by Labour, Matthieu (MLPRO SF) 22 Sep '05

22 Sep '05

Hi Is it possible to have boost function with unlimited arguments For example: boost::function <../boost/function.html> <float (int x, ...)> f; Thank you -------------------------------------------------------- If you are not an intended recipient of this e-mail, please notify the sender, delete it and do not read, act upon, print, disclose, copy, retain or redistribute it. Click here for important additional terms relating to this e-mail. http://www.ml.com/email_terms/ --------------------------------------------------------

3 2

Re: [Boost-users] algorithms and member functions
by Alan M. Carroll 22 Sep '05

22 Sep '05

Using the example in your other e-mail, a solution is find_if(as.begin(), as.end(), boost::bind(std::bitwise_and<int>(), SOME_FLAG, boost::bind(&A::wee, _1))); Anonymous functions and classes are nice. LISP, of course, did it best :-). The closest approximation of this in modern languages is actually in Perl. In Perl, any place you need a code/function pointer, you can write an anonymous function instead of using an existing one. This anonymous function is identical to any other function, it just doesn't have a name (and in fact, you could give it a name later, if you really wanted). It is definitely a very powerful language feature and a lot of the bind / lambda work is intended to support as much of that as possible. I suspect that we're not seeing change in this area because the C++ grammar is already extremely hard to parse and rather cantankerous. Adding this kind of feature might well push it over in to unusablity. A more uniform concept of "code block" as the chunk of stuff enclosed by curly braces would help, but you still have the problem of argument declaration. Perl solves that by having only one argument style so explicit specification is not needed. LISP solved it by embedding the argument declarations inside the code block. C++, unfortunately, puts argument declarations outside the code block while not having a keyword to indicate a function declaration (perhaps, finally, a use for "entry"?). At 11:04 AM 9/21/2005, you wrote: >Hello all, > > >What interests me is the anonymous methods / class style syntax that Java >offers. I've played with it myself, and it seems nice, and one of the >proponents in the afore mentioned flame war advocates it. > >What is the thinking behind not introducing such syntax into C++?

2 1

A question about boost::lambda::if_ behavior
by tom gee 22 Sep '05

22 Sep '05

Can anyone please explain this code: for_each( setFinalVal.begin(), setFinalVal.end(), ( cout<<_1<<" ", if_(_1 < 24.000001 && _1 > 23.999999)[ cout <<"Bingo:<"<< _1<<">"]) ); where setFinalVal is defined as set<double>. I expect it to result in: -32 -24 -16 -12 -11 -10 -9 -8 -7 -5 -4 -3 -2 0 4 5 6 7 8 9 11 12 13 14 24 Bingo:<24>32 40 However, compiled with VC++2005 express, it results in: Bingo:<-32 -24 -16 -12 -11 -10 -9 -8 -7 -5 -4 -3 -2 0 4 5 6 7 8 9 11 12 13 14 24 24>32 40 Why?

1 0