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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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  • Django: Is it possible to attach media files (css, javascript etc) to a View-class?

    - by mrmclovin
    I can't fins any information on how to define css or javascript files in a view like: class MyView(View): .... class Media: css = { 'all' : 'mystyle.css' } If you have a form you can do like: class MyForm(ModelForm): .... class Media: css = { 'all' : 'mystyle.css' } And then in the template you can print the files like; {{ form.media.css }} I like that Syntax very much and I like to keep the View-specific css files in the app-directory. Does anyone know if it's possible? Thanks!

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  • C++ Template Question

    - by user323422
    see following code and please clear doubts1. as ABC is template why it not showing error when we put defination of ABC class member function in test.cpp 2.if i put test.cpp code in test.h , then it working fine // test.h template <typename T> class ABC { public: void foo( T& ); void bar( T& ); }; // test.cpp template <typename T> void ABC<T>::foo( T& ) {} // definition template <typename T> void ABC<T>::bar( T& ) {} // definition template void ABC<char>::foo( char & ); // 1 // main.cpp #include "test.h" int main() { ABC<char> a; a.foo(); // working a.bar(); // link error }

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  • How to know if the argument that is passed to the function is a class, union or enum in c++?

    - by Narek
    I want to define an operator<< for all enums, to cout the value and print that it is an enum like this: code: enum AnyEnum{A,B,C}; AnyEnum enm = A; cout << enm <<endl; output: This is an enum which has a value equal to 0 I know a way of doing this with Boost library by using is_enum struct. But I don’t understand how it works. So that's why, in general, I am interested how to identify if the veriable is a class type, union type or an enum (in compile time).

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  • C++ rvalue temporaries in template

    - by aaa
    hello. Can you please explain me the difference between mechanism of the following: int function(); template<class T> void function2(T&); void main() { function2(function()); // compiler error, instantiated as int & const int& v = function(); function2(v); // okay, instantiated as const int& } is my reasoning correct with respect to instantiation? why is not first instantiated as const T&? Thank you

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  • How to split the definition of template friend funtion within template class?

    - by ~joke
    The following example compiles fine but I can't figure out how to separate declaration and definition of operator<<() is this particular case. Every time I try to split the definition friend is causing trouble and gcc complains the operator<<() definition must take exactly one argument. #include <iostream> template <typename T> class Test { public: Test(const T& value) : value_(value) {} template <typename STREAM> friend STREAM& operator<<(STREAM& os, const Test<T>& rhs) { os << rhs.value_; return os; } private: T value_; }; int main() { std::cout << Test<int>(5) << std::endl; }

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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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  • 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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  • Optimal template for change content via XMLHTTPRequest with JQuery,PHP,SQL [closed]

    - by B.F.
    This is my method to handle XMLHTTPRequests. Avoids mysql request, foreign access, nerves user, double requests. jquery var allow=true; var is_loaded=""; $(document).ready(function(){ .... $(".xx").on("click",functio(){ if(allow){ allow=false; if(is_loaded!="that"){ $.post("job.php", {job:"that",word:"aaa",number:"123"},function(data){ $(".aaa").html(data); is_loaded="that"; }); } setTimeout(function(){allow=true},500); } .... }); job.php <?PHP ob_start('ob_gzhandler'); if(!isset($_SERVER['HTTP_X_REQUESTED_WITH']) or strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) != 'xmlhttprequest')exit("bad boy!"); if($_POST['job']=="that"){ include "includes/that.inc; } elseif($_POST['job']== .... ob_end_flush(); ?> that.inc if(!preg_match("/\w/",$_POST['word'])exit("bad boy!"); if(!is_numeric($_POST['number'])exit("bad boy!"); //exclude more. $path="temp/that_".$row['word']."txt"; if(file_exists($path) and filemtime("includes/that.inc")<$filemtime($path)){ readfile($path); } else{ include "includes/openSql.inc"; $call=sql_query("SELECT * FROM that WHERE name='".mysql_real_escape_string($_POST['word'])."'"); if(!$call)exit("ups"); $out=""; while($row=mysql_fetch_assoc($call)){ $out.=$_POST['word']." loves the color ".$row['color'].".<br/>"; } echo $out; $fn=fopen($path,"wb"); fputs($fn,$out); fclose($fn); } if something change at the database, you just have to delete involved files. Hope it was English.

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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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  • 7 - drupal overriding theme fucntions gettting notices

    - by welovedesign
    So I am overriding a theme function by putting the contents in my template.php module, the problem is that it it throwing up loads of undefined index notices because there are lots of functions that are defined in the module. How can I define these in the template.php file and prevent the notices. Note: I know i can turn them off 'uc_cart_block_content' => array( 'variables' => array( 'help_text' => NULL, 'items' => NULL, 'item_count' => NULL, 'item_text' => NULL, 'total' => NULL, 'summary_links' => NULL, 'collapsed' => TRUE, ), 'file' => 'uc_cart.theme.inc', ),

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  • C++ Template Classes and Copy Construction

    - by themoondothshine
    Is there any way I can construct an new object from the given object if the template parameters of both objects are identical at run-time? For example: I have a template class with the declaration: template<typename _Type1, typename _Type2> class Object; Next, I have two instantiations of the template: template class Object<char, int>; template class Object<wchar_t, wint_t>; Now, I want to write a member function such as: template<typename _Type1, typename _Type2> Object<char, int> Object<_Type1, _Type2>::toCharObject() { if(__gnu_cxx::__are_same<_Type1, char>::__value) return *this; else { //Perform some kind of conversion and return an Object<char, int> } } I have tried a couple of techniques, such as using __gnu_cxx::__enable_if<__gnu_cxx::__are_same<_Type1, char>::__value, _Type1>::__type in a copy constructor for the Oject class, but I keep running into the error: error: conversion from ‘Object<wchar_t, wint_t>’ to non-scalar type ‘Object<char, int>’ requested Is there no way I can do this? Any help will be greatly appreciated!

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  • In a C++ template, is it allowed to return an object with specific type parameters?

    - by nieldw
    When I've got a template with certain type parameters, is it allowed for a function to return an object of this same template, but with different types? In other words, is the following allowed? template<class edgeDecor, class vertexDecor, bool dir> Graph<edgeDecor,int,dir> Graph<edgeDecor,vertexDecor,dir>::Dijkstra(vertex s, bool print = false) const { /* Construct new Graph with apropriate decorators */ Graph<edgeDecor,int,dir> span = new Graph<edgeDecor,int,dir>(); /* ... */ return span; }; If this is not allowed, how can I accomplish the same kind of thing?

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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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  • C++ Template Usage

    - by MalcomTucker
    If I have a template definition like the one below, can someone provide a code sample for how I would actually instantiate an instance of this with two of my own classes? template <class T1, class T2> class LookUpTable { public: LookUpTable(); void set(Tl x, T2* y); T2* get(Tl x); }; Thanks.

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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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  • Template function overloading with identical signatures, why does this work?

    - by user1843978
    Minimal program: #include <stdio.h> #include <type_traits> template<typename S, typename T> int foo(typename T::type s) { return 1; } template<typename S, typename T> int foo(S s) { return 2; } int main(int argc, char* argv[]) { int x = 3; printf("%d\n", foo<int, std::enable_if<true, int>>(x)); return 0; } output: 1 Why doesn't this give a compile error? When the template code is generated, wouldn't the functions int foo(typename T::type search) and int foo(S& search) have the same signature? If you change the template function signatures a little bit, it still works (as I would expect given the example above): template<typename S, typename T> void foo(typename T::type s) { printf("a\n"); } template<typename S, typename T> void foo(S s) { printf("b\n"); } Yet this doesn't and yet the only difference is that one has an int signature and the other is defined by the first template parameter. template<typename T> void foo(typename T::type s) { printf("a\n"); } template<typename T> void foo(int s) { printf("b\n"); } I'm using code similar to this for a project I'm working on and I'm afraid that there's a subtly to the language that I'm not understanding that will cause some undefined behavior in certain cases. I should also mention that it does compile on both Clang and in VS11 so I don't think it's just a compiler bug.

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  • Force type of C++ template

    - by gregseth
    Hi, I've a basic template class, but I'd like to restrain the type of the specialisation to a set of classes or types. e.g.: template <typename T> class MyClass { .../... private: T* _p; }; MyClass<std::string> a; // OK MYCLass<short> b; // OK MyClass<double> c; // not OK Those are just examples, the allowed types may vary. Is that even possible? If it is, how to do so? Thanks.

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