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  • C++ Iterators and inheritance

    - by jomnis
    Have a quick question about what would be the best way to implement iterators in the following: Say I have a templated base class 'List' and two subclasses "ListImpl1" and "ListImpl2". The basic requirement of the base class is to be iterable i.e. I can do: for(List<T>::iterator it = list->begin(); it != list->end(); it++){ ... } I also want to allow iterator addition e.g.: for(List<T>::iterator it = list->begin()+5; it != list->end(); it++){ ... } So the problem is that the implementation of the iterator for ListImpl1 will be different to that for ListImpl2. I got around this by using a wrapper ListIterator containing a pointer to a ListIteratorImpl with subclasses ListIteratorImpl2 and ListIteratorImpl2, but it's all getting pretty messy, especially when you need to implement operator+ in the ListIterator. Any thoughts on a better design to get around these issues?

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  • C++: How to require that one template type is derived from the other

    - by Will
    In a comparison operator: template<class R1, class R2> bool operator==(Manager<R1> m1, Manager<R2> m2) { return m1.internal_field == m2.internal_field; } Is there any way I could enforce that R1 and R2 must have a supertype or subtype relation? That is, I'd like to allow either R1 to be derived from R2, or R2 to be derived from R1, but disallow the comparison if R1 and R2 are unrelated types.

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  • Templated derived class in CRTP (Curiously Recurring Template Pattern)

    - by Butterwaffle
    Hi, I have a use of the CRTP that doesn't compile with g++ 4.2.1, perhaps because the derived class is itself a template? Does anyone know why this doesn't work or, better yet, how to make it work? Sample code and the compiler error are below. Source: foo.C #include <iostream> using namespace std; template<typename X, typename D> struct foo; template<typename X> struct bar : foo<X,bar<X> > { X evaluate() { return static_cast<X>( 5.3 ); } }; template<typename X> struct baz : foo<X,baz<X> > { X evaluate() { return static_cast<X>( "elk" ); } }; template<typename X, typename D> struct foo : D { X operator() () { return static_cast<D*>(this)->evaluate(); } }; template<typename X, typename D> void print_foo( foo<X,D> xyzzx ) { cout << "Foo is " << xyzzx() << "\n"; } int main() { bar<double> br; baz<const char*> bz; print_foo( br ); print_foo( bz ); return 0; } Compiler errors foo.C: In instantiation of ‘foo<double, bar<double> >’: foo.C:8: instantiated from ‘bar<double>’ foo.C:30: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct bar<double>’ foo.C:8: error: declaration of ‘struct bar<double>’ foo.C: In instantiation of ‘foo<const char*, baz<const char*> >’: foo.C:13: instantiated from ‘baz<const char*>’ foo.C:31: instantiated from here foo.C:18: error: invalid use of incomplete type ‘struct baz<const char*>’ foo.C:13: error: declaration of ‘struct baz<const char*>’

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  • Obtain container type from (its) iterator type in C++ (STL)

    - by KRao
    It is easy given a container to get the associated iterators, example: std::vector<double>::iterator i; //An iterator to a std::vector<double> I was wondering if it is possible, given an iterator type, to deduce the type of the "corresponding container" (here I am assuming that for each container there is one and only one (non-const) iterator). More precisely, I would like a template metafunction that works with all STL containers (without having to specialize it manually for each single container) such that, for example: ContainerOf< std::vector<double>::iterator >::type evaluates to std::vector<double> Is it possible? If not, why? Thank you in advance for any help!

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  • Why can't I create a templated sublcass of System::Collections::Generic::IEnumerable<T>?

    - by fiirhok
    I want to create a generic IEnumerable implementation, to make it easier to wrap some native C++ classes. When I try to create the implementation using a template parameter as the parameter to IEnumerable, I get an error. Here's a simple version of what I came up with that demonstrates my problem: ref class A {}; template<class B> ref class Test : public System::Collections::Generic::IEnumerable<B^> // error C3225... {}; void test() { Test<A> ^a = gcnew Test<A>(); } On the indicated line, I get this error: error C3225: generic type argument for 'T' cannot be 'B ^', it must be a value type or a handle to a reference type If I use a different parent class, I don't see the problem: template<class P> ref class Parent {}; ref class A {}; template<class B> ref class Test : public Parent<B^> // no problem here {}; void test() { Test<A> ^a = gcnew Test<A>(); } I can work around it by adding another template parameter to the implementation type: ref class A {}; template<class B, class Enumerable> ref class Test : public Enumerable {}; void test() { using namespace System::Collections::Generic; Test<A, IEnumerable<A^>> ^a = gcnew Test<A, IEnumerable<A^>>(); } But this seems messy to me. Also, I'd just like to understand what's going on here - why doesn't the first way work?

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  • specyfic syntax question

    - by bua
    Hi there, Is it possible to create template to the initialization like: template <typename C> typename C::value_type fooFunction(C& c) {...}; std::vector<string> vec_instance; fooFunction(cont<0>(vec_instance)); fooFunction(cont<1>(vec_instance)); In general i'm interested is it possible to specify template using integer (ie. 0) instead of true type name. And how to achieve above?

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  • C++, what does this syntax mean?

    - by aaa
    i found this in this file: http://www.boost.org/doc/libs/1_43_0/boost/spirit/home/phoenix/core/actor.hpp What does this syntax means? struct actor ... { ... template <typename T0, typename T1> typename result<actor(T0&,T1&)>::type // this line thank you

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  • Magento CSS not Loading in New Template

    - by vulgarbulgar
    I have posted on the template creator's site, as well as Magento support, but no one has responded. The CSS is not loading at all on the custom theme I have installed, which is supposedly compatible with the current version of Magento. You can view the page here: shop.dearearth.net This is a fresh installation of Magento and the theme, with a fresh database. I have a feeling it should be a relatively quick fix. Thanks for looking.

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  • Django template CSS/IMG is "off" in the URL

    - by erimar77
    I have /path/to/my/theme/static/css/frontend.css which is called by base.html <link rel="stylesheet" type="text/css" href="{{ STATIC_URL }}css/frontend.css" media="all" /> In which I've got a background for the header: #header-wrapper min-width: 960px; height: 150px; background: transparent url(img/header-bg.png) repeat-x center bottom; } The file is /path/to/my/theme/static/img I've run manage.py collectstatic to gather the files and almost everything looks correct except the link generated looks like: http://example.com/static/css/img/header-bg.png In which the image does not show, because the correct URL is: http://example.com/static/img/header-bg.png Where am I going wrong??

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  • Usage of CRTP in a call chain

    - by fhw72
    In my widget library I'd like to implement some kind of call chain to initialize a user supplied VIEW class which might(!) be derived from another class which adds some additional functionality like this: #include <iostream> template<typename VIEW> struct App { VIEW view; void init() {view.initialize(); } }; template<typename DERIVED> struct SpecializedView { void initialize() { std::cout << "SpecializedView" << std::endl; static_cast<DERIVED*>(this)->initialize(); } }; struct UserView : SpecializedView<UserView> { void initialize() {std::cout << "UserView" << std::endl; } }; int _tmain(int argc, _TCHAR* argv[]) { // Cannot be altered to: App<SpecializedView<UserView> > app; App<UserView> app; app.init(); return 0; } Is it possible to achieve some kind of call chain (if the user supplied VIEW class is derived from "SpecializedView") such that the output will be: console output: SpecializedView UserView Of course it would be easy to instantiate variable app with the type derived from but this code is hidden in the library and should not be alterable. In other words: The library code should only get the user derived type as parameter.

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  • Class templating std::set key types

    - by TomFLuff
    I have a class to evaluate set algebra but wish to template it. At the minute it looks a bit like this set.h: template<typename T> class SetEvaluation { public: SetEvaluation<T>(); std::set<T> evaluate(std::string in_expression); } set.cpp template<typename T> std::set<T> SetEvaluation<T>::evaluate(std::string expression) { std::set<T> result; etc etc... } But i'm getting undefined reference errors when compiling. Is it possible to declare the return type as std::set<T> and then pass std::string as the class template param. There are no errors in the class but only when I try to instantiate SetEvaluation<std::string> Can anyone shed light on this problem? thanks

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  • Django - problem with {% url facebook_xd_receiver %}

    - by Gaurav
    I'm using {% url facebook_xd_receiver %} in one of my HTML files. This works just fine when I run my project using the command python manage.py runserver But the same project stops running and gives me a "TemplateSyntaxError" at the line {% url facebook_xd_receiver %} Can anyone please tell me what could be the difference between the dev server run through the command line and the apache server. Is there anything I'm missing out on while configuring the Apache server? Or is it a Django problem?

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  • Is it possible to supply template parameters when calling operator()?

    - by Paul
    I'd like to use a template operator() but am not sure if it's possible. Here is a simple test case that won't compile. Is there something wrong with my syntax, or is this simply not possible? struct A { template<typename T> void f() { } template<typename T> void operator()() { } }; int main() { A a; a.f<int>(); // This compiles. a.operator()<int>(); // This compiles. a<int>(); // This won't compile. return 0; }

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  • templated class : accessing derived normal-class methods

    - by user1019129
    I have something like this : class Container1 { public: method1() { ... } } class Container2 { public: method1() { ... } } template<class C = Container1> class X : public C { public: using C::method1(); ..... X(string& str) : C(str) {}; X& other_method() { method1(); ...; } } My question is why I have to use "using C::method1()", to be able to access the method.. Most of answers I found is for the case where templated-class inhering templated-class. Normally they mention using "this-", but this does not seem to work in this case. Can I do something else shorter... Also I'm suspecting the other error I'm getting is related to the same problem : no match call for (X<Container1>) (<std::string&>)

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  • How to loop X times in Django?

    - by Mark
    I have user reviews on my site. Each review has a rating of 1-5 stars. I want to print that many stars. How do I do it? I only see {% for X in Y %} which lets you iterate over a list, but not a certain number of times.

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  • Template specialization to use default type if class member typedef does not exist

    - by Frank
    Hi Everyone, I'm trying to write code that uses a member typedef of a template argument, but want to supply a default type if the template argument does not have that typedef. A simplified example I've tried is this: struct DefaultType { DefaultType() { printf("Default "); } }; struct NonDefaultType { NonDefaultType() { printf("NonDefault "); } }; struct A {}; struct B { typedef NonDefaultType Type; }; template<typename T, typename Enable = void> struct Get_Type { typedef DefaultType Type; }; template<typename T> struct Get_Type< T, typename T::Type > { typedef typename T::Type Type; }; int main() { Get_Type::Type test1; Get_Type::Type test2; } I would expect this to print "Default NonDefault", but instead it prints "Default Default". My expectation is that the second line in main() should match the specialized version of Get_Type, because B::Type exists. However, this does not happen. Can anyone explain what's going on here and how to fix it, or another way to accomplish the same goal? Thank you.

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  • C++ template restrictions

    - by pingvinus
    I wondering is there any way to set restrictions on template class? Specify that every type substituted in template must have specific ancestor (realize some interface). template < class B > //and every B must be a child of abstract C class A { public: B * obj; int f() { return B::x + this->obj->f(); } }; Like = in haskell func :: (Ord a, Show b) => a -> b -> c

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