Search Results

Search found 2666 results on 107 pages for 'boost lambda'.

Page 25/107 | < Previous Page | 21 22 23 24 25 26 27 28 29 30 31 32  | Next Page >

  • in sending with name using ? (counting), how does the process happen? [closed]

    - by sam
    Sometimes expressions has two states. the result of analyses might be two different things but at the end, there might be the same answer(result) for them.(the way we get to the answer is different,but the result is the same) for example: P-Q = M P-T = M in lambda we have two solutions. 1.sending with name. (in ALGOL it have over load) 2.sending with value Example: Q: (?y.(yy) (?x.(xx)a)) 1.sending with name (?x.(xx)a ?x.(xx)a) ((aa)(aa)) 2.sending with value (?y.(yy) aa) ((aa)(aa)) now here is the question, in sending with name that used ? (counting), how does the process happen? How this sending (transmittal) happen? How does it work?

    Read the article

  • Containers of reference_wrappers (comparison operators required?)

    - by kloffy
    If you use stl containers together with reference_wrappers of POD types, the following code works just fine: int i = 3; std::vector< boost::reference_wrapper<int> > is; is.push_back(boost::ref(i)); std::cout << (std::find(is.begin(),is.end(),i)!=is.end()) << std::endl; However, if you use non-POD types such as (contrived example): struct Integer { int value; bool operator==(const Integer& rhs) const { return value==rhs.value; } bool operator!=(const Integer& rhs) const { return !(*this == rhs); } }; It doesn't suffice to declare those comparison operators, instead you have to declare: bool operator==(const boost::reference_wrapper<Integer>& lhs, const Integer& rhs) { return boost::unwrap_ref(lhs)==rhs; } And possibly also: bool operator==(const Integer& lhs, const boost::reference_wrapper<Integer>& rhs) { return lhs==boost::unwrap_ref(rhs); } In order to get the equivalent code to work: Integer j = { 0 }; std::vector< boost::reference_wrapper<Integer> > js; js.push_back(boost::ref(j)); std::cout << (std::find(js.begin(),js.end(),j)!=js.end()) << std::endl; Now, I'm wondering if this is really the way it's meant to be done, since it seems impractical. It just seems there should be a simpler solution, e.g. templates: template<class T> bool operator==(const boost::reference_wrapper<T>& lhs, const T& rhs) { return boost::unwrap_ref(lhs)==rhs; } template<class T> bool operator==(const T& lhs, const boost::reference_wrapper<T>& rhs) { return lhs==boost::unwrap_ref(rhs); } There's probably a good reason why reference_wrapper behaves the way it does (possibly to accomodate non-POD types without comparison operators?). Maybe there already is an elegant solution and I just haven't found it.

    Read the article

  • Using time facets on universal_time

    - by scooterman
    Hi all, on boost, to create a time facet to format an specified time we use the folowing: boost::local_time::local_time_facet* facet = new boost::local_time::local_time_facet("%Y%m%d %H:%M:%S.%f"); std::stringstream date_stream; date_stream.imbue(std::locale(date_stream.getloc(), facet)); date_stream << boost::local_time::local_microsec_clock::local_time(boost::local_time::time_zone_ptr()); How do I do the same thing, but using an universal clock: boost::posix_time::microsec_clock::universal_time() Thanks

    Read the article

  • how a thread can signal when it's finished?

    - by Kyle
    #include <iostream> #include <boost/thread.hpp> using std::endl; using std::cout; using namespace boost; mutex running_mutex; struct dostuff { volatile bool running; dostuff() : running(true) {} void operator()(int x) { cout << "dostuff beginning " << x << endl; this_thread::sleep(posix_time::seconds(2)); cout << "dostuff is done doing stuff" << endl; mutex::scoped_lock running_lock(running_mutex); running = false; } }; bool is_running(dostuff& doer) { mutex::scoped_lock running_lock(running_mutex); return doer.running; } int main() { cout << "Begin.." << endl; dostuff doer; thread t(doer, 4); if (is_running(doer)) cout << "Cool, it's running.\n"; this_thread::sleep(posix_time::seconds(3)); if (!is_running(doer)) cout << "Cool, it's done now.\n"; else cout << "still running? why\n"; // This happens! :( return 0; } Why is the output of the above program: Begin.. Cool, it's running. dostuff beginning 4 dostuff is done doing stuff still running? why How can dostuff correctly flag when it is done? I do not want to sit around waiting for it, I just want to be notified when it's done.

    Read the article

  • Best boost to productivity : SSD or second screen ?

    - by jfoucher
    Within the same budget, which do you think would be the best boost to productivity for general web development : An SSD as boot drive, or a second screen ? I currently use a 16.4" laptop with full HD screen and 7200 RPM HDD Thanks EDIT: In light of the answers so far, I add that I work at home and while I have a home office with a proper desk, I'm just as often coding sitting on the sofa in the living room.

    Read the article

  • 5 Easy Ways to Get High PR Links to Boost Your Site in Google

    High PR links are some of the most valuable aspects in any link-building & SEO campaign. Not only do these links make Google respect your site more, but they can also boost your site's ranking overnight. Here are 5 places to get high quality links that will do a lot of good to your site's ranking in Google.

    Read the article

  • How SEO Services by a SEO Company Can Boost Your Sales

    If we look at the present scenario the importance of growing your business and expanding your online brand recognition using all the strategic SEO elements available can just not be overstated. Today to be the very best at marketing the business or even the websites has to reach to its potential customers and hiring SEO companies and SEO experts is proving best method to keep track of the latest developments in search engine optimization. In this article, know how taking help of SEO services from any SEO company can actually boost up your sales.

    Read the article

  • How SEO Services by a SEO Company Can Boost Your Sales

    If we look at the present scenario the importance of growing your business and expanding your online brand recognition using all the strategic SEO elements available can just not be overstated. Today to be the very best at marketing the business or even the websites has to reach to its potential customers and hiring SEO companies and SEO experts is proving best method to keep track of the latest developments in search engine optimization. In this article, know how taking help of SEO services from any SEO company can actually boost up your sales.

    Read the article

  • Advanced SEO Strategy Guaranteed to Boost Your SEO

    One of the main factors for optimising your website for page 1 on Google is to get incoming links from other web pages, and to boost the effectiveness of those links there are a few things to consider, including the relevance of the page where your link is coming from, plus the link itself needs to be relevant to the keywords that you are targeting on Google. Here's one strategy you can use for getting great incoming links to help you achieve higher Google rankings.

    Read the article

  • Boost Profits on Your Website

    By now you may have already built a website but are still not raking in profits from it. Your next step is to use some SEO strategies to boost its search rankings and convince site visitors to either buy something from you or avail of an offer.

    Read the article

  • Forum Sites - Boost Your Search Marketing

    Forum sites boost your search marketing. Why? Finding the right site for your business is essential to boosting your online marketing; find the right community, one that deals with your specific services. Don't waste your time in a site that has no relevance to what you're offering. Find out the hub sites for your services by using keywords related to you.

    Read the article

  • How does versioning work when using Boost Serialization for Derived Classes?

    - by Venkata Adusumilli
    When a Client serializes the following data: InternationalStudent student; student.id("Client ID"); student.firstName("Client First Name"); student.country("Client Country"); the Server receives the following: ID = "Client ID" Country = "Client First Name" instead of the following: ID = "Client ID" Country = "Client Country" The only difference between the Server and Client classes is the First Name of the Student. How can we make the Server ignore First Name recieved from the Client and process the Country? Server Side Classes class Student { public: Student(){} virtual ~Student(){} public: std::string id() { return idM; } void id(std::string id) { idM = id; } protected: friend class boost::serialization::access; protected: std::string idM; protected: template<class A> void serialize(A& archive, const unsigned int /*version*/) { archive & BOOST_SERIALIZATION_NVP(idM); } }; class InternationalStudent : public Student { public: InternationalStudent() {} ~InternationalStudent() {} public: std::string country() { return countryM; } void country(std::string country) { countryM = country; } protected: friend class boost::serialization::access; protected: std::string countryM; protected: template<class A> void serialize(A& archive, const unsigned int /*version*/) { archive & BOOST_SERIALIZATION_NVP(boost::serialization::base_object<Student>(*this)); archive & BOOST_SERIALIZATION_NVP(countryM); } }; Client Side Classes class Student { public: Student(){} virtual ~Student(){} public: std::string id() { return idM; } void id(std::string id) { idM = id; } std::string firstName() { return firstNameM; } void firstName(std::string name) { firstNameM = name; } protected: friend class boost::serialization::access; protected: std::string idM; std::string firstNameM; protected: template<class A> void serialize(A& archive, const unsigned int /*version*/) { archive & BOOST_SERIALIZATION_NVP(idM); if (version >=1) { archive & BOOST_SERIALIZATION_NVP(firstNameM); } } }; BOOST_CLASS_VERSION(Student, 1) class InternationalStudent : public Student { public: InternationalStudent() {} ~InternationalStudent() {} public: std::string country() { return countryM; } void country(std::string country) { countryM = country; } protected: friend class boost::serialization::access; protected: std::string countryM; protected: template<class A> void serialize(A& archive, const unsigned int /*version*/) { archive & BOOST_SERIALIZATION_NVP(boost::serialization::base_object<Student>(*this)); archive & BOOST_SERIALIZATION_NVP(countryM); } };

    Read the article

  • How can I implement a database TableView like thing in C++?

    - by Industrial-antidepressant
    How can I implement a TableView like thing in C++? I want to emulating a tiny relation database like thing in C++. I have data tables, and I want to transform it somehow, so I need a TableView like class. I want filtering, sorting, freely add and remove items and transforming (ex. view as UPPERCASE and so on). The whole thing is inside a GUI application, so datatables and views are attached to a GUI (or HTML or something). So how can I identify an item in the view? How can I signal it when the table is changed? Is there some design pattern for this? Here is a simple table, and a simple data item: #include <string> #include <boost/multi_index_container.hpp> #include <boost/multi_index/member.hpp> #include <boost/multi_index/ordered_index.hpp> #include <boost/multi_index/random_access_index.hpp> using boost::multi_index_container; using namespace boost::multi_index; struct Data { Data() {} int id; std::string name; }; struct row{}; struct id{}; struct name{}; typedef boost::multi_index_container< Data, indexed_by< random_access<tag<row> >, ordered_unique<tag<id>, member<Data, int, &Data::id> >, ordered_unique<tag<name>, member<Data, std::string, &Data::name> > > > TDataTable; class DataTable { public: typedef Data item_type; typedef TDataTable::value_type value_type; typedef TDataTable::const_reference const_reference; typedef TDataTable::index<row>::type TRowIndex; typedef TDataTable::index<id>::type TIdIndex; typedef TDataTable::index<name>::type TNameIndex; typedef TRowIndex::iterator iterator; DataTable() : row_index(rule_table.get<row>()), id_index(rule_table.get<id>()), name_index(rule_table.get<name>()), row_index_writeable(rule_table.get<row>()) { } TDataTable::const_reference operator[](TDataTable::size_type n) const { return rule_table[n]; } std::pair<iterator,bool> push_back(const value_type& x) { return row_index_writeable.push_back(x); } iterator erase(iterator position) { return row_index_writeable.erase(position); } bool replace(iterator position,const value_type& x) { return row_index_writeable.replace(position, x); } template<typename InputIterator> void rearrange(InputIterator first) { return row_index_writeable.rearrange(first); } void print_table() const; unsigned size() const { return row_index.size(); } TDataTable rule_table; const TRowIndex& row_index; const TIdIndex& id_index; const TNameIndex& name_index; private: TRowIndex& row_index_writeable; }; class DataTableView { DataTableView(const DataTable& source_table) {} // How can I implement this? // I want filtering, sorting, signaling upper GUI layer, and sorting, and ... }; int main() { Data data1; data1.id = 1; data1.name = "name1"; Data data2; data2.id = 2; data2.name = "name2"; DataTable table; table.push_back(data1); DataTable::iterator it1 = table.row_index.iterator_to(table[0]); table.erase(it1); table.push_back(data1); Data new_data(table[0]); new_data.name = "new_name"; table.replace(table.row_index.iterator_to(table[0]), new_data); for (unsigned i = 0; i < table.size(); ++i) std::cout << table[i].name << std::endl; #if 0 // using scenarios: DataTableView table_view(table); table_view.fill_from_source(); // synchronization with source table_view.remove(data_item1); // remove item from view table_view.add(data_item2); // add item from source table table_view.filter(filterfunc); // filtering table_view.sort(sortfunc); // sorting // modifying from source_able, hot to signal the table_view? // FYI: Table view is atteched to a GUI item table.erase(data); table.replace(data); #endif return 0; }

    Read the article

  • Using in the same time Boost in release and debug mode

    - by Oodini
    Hello, The title is just for teasing. :-) I know it isn't possible, but here is my problem. My app (a DLL, actually) uses Boost. It also uses an external API, which also uses Boost. The external API is of course provided in a release binary, anlong the needed release Boost binaries. When I compile/link my DLL in release mode, I have no problem. I precise I link my app to Boost in a dynamic way (BOOST_ALL_DYN_LINK). In debug mode, I can't load my DLL. I am not sure it is because of Boost, but I suspect Windows doesn't like working with two Boost (the release one called by the external lib, and the debug one called by my code). Will it work better if I link my code statically with the release Boost ? I don't think it is related to CRT, because I have nothing in the Events Viewer. I use Visual Studio 2005 SP1. Thanks.

    Read the article

  • Equvalent c++0x program withought using boost threads..

    - by Eternal Learner
    I have the below simple program using boost threads, what would be the changes needed to do the same in c++0X #include<iostream> #include<boost/thread/thread.hpp> boost::mutex mutex; struct count { count(int i): id(i){} void operator()() { boost::mutex::scoped_lock lk(mutex); for(int i = 0 ; i < 10000 ; i++) { std::cout<<"Thread "<<id<<"has been called "<<i<<" Times"<<std::endl; } } private: int id; }; int main() { boost::thread thr1(count(1)); boost::thread thr2(count(2)); boost::thread thr3(count(3)); thr1.join(); thr2.join(); thr3.join(); return 0; }

    Read the article

  • Mutating the expression tree of a predicate to target another type

    - by Jon
    Intro In the application I 'm currently working on, there are two kinds of each business object: the "ActiveRecord" type, and the "DataContract" type. So for example, we have: namespace ActiveRecord { class Widget { public int Id { get; set; } } } namespace DataContracts { class Widget { public int Id { get; set; } } } The database access layer takes care of "translating" between hierarchies: you can tell it to update a DataContracts.Widget, and it will magically create an ActiveRecord.Widget with the same property values and save that. The problem I have surfaced when attempting to refactor this database access layer. The Problem I want to add methods like the following to the database access layer: // Widget is DataContract.Widget interface DbAccessLayer { IEnumerable<Widget> GetMany(Expression<Func<Widget, bool>> predicate); } The above is a simple general-use "get" method with custom predicate. The only point of interest is that I 'm not passing in an anonymous function but rather an expression tree. This is done because inside DbAccessLayer we have to query ActiveRecord.Widget efficiently (LINQ to SQL) and not have the database return all ActiveRecord.Widget instances and then filter the enumerable collection. We need to pass in an expression tree, so we ask for one as the parameter for GetMany. The snag: the parameter we have needs to be magically transformed from an Expression<Func<DataContract.Widget, bool>> to an Expression<Func<ActiveRecord.Widget, bool>>. This is where I haven't managed to pull it off... Attempted Solution What we 'd like to do inside GetMany is: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( predicate.Body, predicate.Parameters); // use lambda to query ActiveRecord.Widget and return some value } This won't work because in a typical scenario, for example if: predicate == w => w.Id == 0; ...the expression tree contains a MemberAccessExpression instance which has a MemberInfo property (named Member) that point to members of DataContract.Widget. There are also ParameterExpression instances both in the expression tree and in its parameter expression collection (predicate.Parameters); After searching a bit, I found System.Linq.Expressions.ExpressionVisitor (its source can be found here in the context of a how-to, very helpful) which is a convenient way to modify an expression tree. Armed with this, I implemented a visitor. This simple visitor only takes care of changing the types in member access and parameter expressions. It may not be complete, but it's fine for the expression w => w.Id == 0. internal class Visitor : ExpressionVisitor { private readonly Func<Type, Type> dataContractToActiveRecordTypeConverter; public Visitor(Func<Type, Type> dataContractToActiveRecordTypeConverter) { this.dataContractToActiveRecordTypeConverter = dataContractToActiveRecordTypeConverter; } protected override Expression VisitMember(MemberExpression node) { var dataContractType = node.Member.ReflectedType; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); var converted = Expression.MakeMemberAccess( base.Visit(node.Expression), activeRecordType.GetProperty(node.Member.Name)); return converted; } protected override Expression VisitParameter(ParameterExpression node) { var dataContractType = node.Type; var activeRecordType = this.dataContractToActiveRecordTypeConverter(dataContractType); return Expression.Parameter(activeRecordType, node.Name); } } With this visitor, GetMany becomes: IEnumerable<DataContract.Widget> GetMany( Expression<Func<DataContract.Widget, bool>> predicate) { var visitor = new Visitor(...); var lambda = Expression.Lambda<Func<ActiveRecord.Widget, bool>>( visitor.Visit(predicate.Body), predicate.Parameters.Select(p => visitor.Visit(p)); var widgets = ActiveRecord.Widget.Repository().Where(lambda); // This is just for reference, see below Expression<Func<ActiveRecord.Widget, bool>> referenceLambda = w => w.Id == 0; // Here we 'd convert the widgets to instances of DataContract.Widget and // return them -- this has nothing to do with the question though. } Results The good news is that lambda is constructed just fine. The bad news is that it isn't working; it's blowing up on me when I try to use it (the exception messages are really not helpful at all). I have examined the lambda my code produces and a hardcoded lambda with the same expression; they look exactly the same. I spent hours in the debugger trying to find some difference, but I can't. When predicate is w => w.Id == 0, lambda looks exactly like referenceLambda. But the latter works with e.g. IQueryable<T>.Where, while the former does not (I have tried this in the immediate window of the debugger). I should also mention that when predicate is w => true, it all works just fine. Therefore I am assuming that I 'm not doing enough work in Visitor, but I can't find any more leads to follow on. Can someone point me in the right direction? Thanks in advance for your help!

    Read the article

  • C++ Multithreaded server help

    - by kisplit
    Hello all, I'm working on a multithreaded server in c++ using boost-asio. Currently a design problem I'm running into deals with erasing a connection. I have a single server instance which holds a vector of connection objects. These connections receive commands which I parse. One command in particular deals with sending data to ALL connections in my vector. Now when a connection disconnects I'm currently erasing this connection from the vector and calling the destructor. It seems like I'm going to run into problems when someone 'SendAll' at the same time someone 'Disconnect'. Could anyone recommend a better design or just point me in the right direction? Any help greatly appreciated. Thanks

    Read the article

  • What's correct way to remove a boost::shared_ptr from a list?

    - by Catskul
    I have a std::list of boost::shared_ptr<T> and I want to remove an item from it but I only have a pointer of type T* which matches one of the items in the list. However I cant use myList.remove( tPtr ) I'm guessing because shared_ptr does not implement == for its template argument type. My immediate thought was to try myList.remove( shared_ptr<T>(tPtr) ) which is syntactically correct but it will crash from a double delete since the temporary shared_ptr has a separate use_count. std::list< boost::shared_ptr<T> > myList; T* tThisPtr = new T(); // This is wrong; only done for example code. // stand-in for actual code in T using // T's actual "this" pointer from within T { boost::shared_ptr<T> toAdd( tThisPtr ); // typically would be new T() myList.push_back( toAdd ); } { //T has pointer to myList so that upon a certain action, // it will remove itself romt the list //myList.remove( tThisPtr); //doesn't compile myList.remove( boost::shared_ptr<T>(tThisPtr) ); // compiles, but causes // double delete } The only options I see remaining are to use std::find with a custom compare, or to loop through the list brute force and find it myself, but it seems there should be a better way. Am I missing something obvious, or is this just too non-standard a use to be doing a remove the clean/normal way?

    Read the article

  • Is it possible to use boost::bind to effectively concatenate functions?

    - by Catskul
    Assume that I have a boost::function of with an arbitrary signature called type CallbackType. Is it possible to use boost::bind to compose a function that takes the same arguments as the CallbackType but calls the two functors in succession? Hypothetical example using a magic template: Template<typename CallbackType> class MyClass { public: CallbackType doBoth; MyClass( CallbackType callback ) { doBoth = bind( magic<CallbackType>, protect( bind(&MyClass::alert, this) ), protect( callback ) ); } void alert() { cout << "It has been called\n"; } }; void doIt( int a, int b, int c) { cout << "Doing it!" << a << b << c << "\n"; } int main() { typedef boost::function<void (int, int, int)> CallbackType; MyClass<CallbackType> object( boost::bind(doIt) ); object.doBoth(); return 0; }

    Read the article

  • boost::unordered_map is... ordered?

    - by Thanatos
    I have a boost::unordered_map, but it appears to be in order, giving me an overwhelming feeling of "You're Doing It Wrong". Why is the output to this in order? I would've expected the underlying hashing algorithm to have randomized this order: #include <iostream> #include <boost/unordered_map.hpp> int main() { boost::unordered_map<int, int> im; for(int i = 0; i < 50; ++i) { im.insert(std::make_pair(i, i)); } boost::unordered_map<int, int>::const_iterator i; for(i = im.begin(); i != im.end(); ++i) { std::cout << i->first << ", " << i->second << std::endl; } return 0; } ...gives me... 0, 0 1, 1 2, 2 ... 47, 47 48, 48 49, 49

    Read the article

< Previous Page | 21 22 23 24 25 26 27 28 29 30 31 32  | Next Page >