Use is_same: #include <type_traits> template <typename T> void foo() { if (std::is_same<T, animal>::value) { /* … */ } // optimizable… } Usually, that’s a totally unworkable design, though, and you really want to specialize: template <typename T> void foo() { /* generic implementation */ } template <> void foo<animal>() { /* specific for T = … Read more
You should be able to do something like this in your respond_to block: respond_to do |format| format.json render :partial => “users/show.json” end which will render the template in app/views/users/_show.json.erb.
The problem is not that you can’t forward-declare a template class. Yes, you do need to know all of the template parameters and their defaults to be able to forward-declare it correctly: namespace std { template<class T, class Allocator = std::allocator<T>> class list; } But to make even such a forward declaration in namespace std … Read more
For understanding templates, it’s of huge advantage to get the terminology straight because the way you speak about them determines the way to think about them. Specifically, Area is not a template class, but a class template. That is, it is a template from which classes can be generated. Area<int> is such a class (it’s … Read more
I’ve done a turing machine in C++11. Features that C++11 adds are not significant for the turing machine indeed. It just provides for arbitrary length rule lists using variadic templates, instead of using perverse macro metaprogramming :). The names for the conditions are used to output a diagram on stdout. i’ve removed that code to … Read more
If you want the enum names themselves as strings, see this post. Otherwise, a std::map<MyEnum, char const*> will work nicely. (No point in copying your string literals to std::strings in the map) For extra syntactic sugar, here’s how to write a map_init class. The goal is to allow std::map<MyEnum, const char*> MyMap; map_init(MyMap) (eValue1, “A”) … Read more
Unlike an unscoped enumeration, a scoped enumeration is not implicitly convertible to its integer value. You need to explicitly convert it to an integer using a cast: std::cout << static_cast<std::underlying_type<A>::type>(a) << std::endl; You may want to encapsulate the logic into a function template: template <typename Enumeration> auto as_integer(Enumeration const value) -> typename std::underlying_type<Enumeration>::type { return … Read more
ng-container is preferred over template: <ng-container *ngIf=”expression”> See: Angular 2 ng-container https://github.com/angular/angular.io/issues/2303
Yes, it is a non-type parameter. You can have several kinds of template parameters Type Parameters. Types Templates (only classes and alias templates, no functions or variable templates) Non-type Parameters Pointers References Integral constant expressions What you have there is of the last kind. It’s a compile time constant (so-called constant expression) and is of … Read more
There is no way to explicitly specify the template arguments when calling a constructor template, so they have to be deduced through argument deduction. This is because if you say: Foo<int> f = Foo<int>(); The <int> is the template argument list for the type Foo, not for its constructor. There’s nowhere for the constructor template’s … Read more