There are a few ways to echo HTML in PHP. 1. In between PHP tags <?php if(condition){ ?> <!– HTML here –> <?php } ?> 2. In an echo if(condition){ echo “HTML here”; } With echos, if you wish to use double quotes in your HTML you must use single quote echos like so: echo … Read more
I made this short example which also works. #include <iostream> #include <type_traits> class foo; class bar; template<class T> struct is_bar { template<class Q = T> typename std::enable_if<std::is_same<Q, bar>::value, bool>::type check() { return true; } template<class Q = T> typename std::enable_if<!std::is_same<Q, bar>::value, bool>::type check() { return false; } }; int main() { is_bar<foo> foo_is_bar; is_bar<bar> bar_is_bar; … Read more
The compiler needs to have access to the entire template definition (not just the signature) in order to generate code for each instantiation of the template, so you need to move the definitions of the functions to your header. For more details read about The Inclusion Model.
From C++ Templates The Complete Guide: Member function templates cannot be declared virtual. This constraint is imposed because the usual implementation of the virtual function call mechanism uses a fixed-size table with one entry per virtual function. However, the number of instantiations of a member function template is not fixed until the entire program has … Read more
Write a custom template filter: from django.template.defaulttags import register … @register.filter def get_item(dictionary, key): return dictionary.get(key) (I use .get so that if the key is absent, it returns none. If you do dictionary[key] it will raise a KeyError then.) usage: {{ mydict|get_item:item.NAME }}
I think you need to use template template syntax to pass a parameter whose type is a template dependent on another template like this: template <template<class> class H, class S> void f(const H<S> &value) { } Here, H is a template, but I wanted this function to deal with all specializations of H. NOTE: I’ve … Read more
Here’s a possible implementation relying on C++11 features. It correctly detects the function even if it’s inherited (unlike the solution in the accepted answer, as Mike Kinghan observes in his answer). The function this snippet tests for is called serialize: #include <type_traits> // Primary template with a static assertion // for a meaningful error message … Read more
This is one of those frequently asked questions that have different approaches that are similar but not really the same. The three approaches differ in who you are declaring to be a friend of your function –and then on how you implement it. The extrovert Declare all instantiations of the template as friends. This is … Read more
You can use this-> to make clear that you are referring to a member of the class: void Bar<T>::BarFunc () { std::cout << this->_foo_arg << std::endl; } Alternatively you can also use “using” in the method: void Bar<T>::BarFunc () { using Bar<T>::_foo_arg; // Might not work in g++, IIRC std::cout << _foo_arg << std::endl; } … Read more
Deduction refers to the process of determining the type of a template parameter from a given argument. It applies to function templates, auto, and a few other cases (e.g. partial specialization). For example, consider: template <typename T> void f(std::vector<T>); Now if you say f(x), where you declared std::vector<int> x;, then T is deduced as int, … Read more