This is due to a really obscure provision of the standard in which if you have a template that tries to access a template function in an object whose type depends on a template argument, you have to use the template keyword in a weird way: this->template f<int>(); This is similar to the weirdness with … Read more
Why the loop? You could simply do this: {% if ‘priority’ in data %} <p>Priority: {{ data[‘priority’] }}</p> {% endif %} When you were originally doing your string comparison, you should have used == instead.
Another approach, which is slightly more generic than Ben’s, is as follows: #include <tuple> template <typename… Args> struct variadic_typedef { // this single type represents a collection of types, // as the template arguments it took to define it }; template <typename… Args> struct convert_in_tuple { // base case, nothing special, // just use the … Read more
A string literal cannot be used as a template argument. Update: Nowadays, a few years after this question was asked and answered, it is possible to use string literals as template arguments. With C++11, we can use characters packs as template arguments (template<char …c>) and it is possible to pass a literal string to such … Read more
How is std::swap implemented? Yes, the implementation presented in the question is the classic C++03 one. A more modern (C++11) implementation of std::swap looks like this: template<typename T> void swap(T& t1, T& t2) { T temp = std::move(t1); // or T temp(std::move(t1)); t1 = std::move(t2); t2 = std::move(temp); } This is an improvement over the … Read more
The problem is you can’t template a typedef, also there is no need to typedef structs in C++. The following will do what you need template <typename T> struct array { size_t x; T *ary; };
derived is incomplete when you use it as a template argument to base in its base classes list. A common workaround is to use a traits class template. Here’s your example, traitsified. This shows how you can use both types and functions from the derived class through the traits. // Declare a base_traits traits class … Read more
There is no built-in mechanism for this. typeid(T)::name() can give some info, but the standard does not mandate this string to be human-readable; just that it has to be distinct for each type. (E.x. Microsoft Visual C++ uses human-readable strings; GCC does not.) You can build your own system though. For example, traits-based. Something like … Read more
The compiler told you exactly what to do. Write typename before ptrModel<std::vector<Data> >::Type, like so: typedef typename ptrModel<std::vector<Data> >::Type Type; The reason for this requirement is that the compiler doesn’t at this point know whether ptrModel<std::vector<Data> >::Type describes a member variable or a nested type. It can’t even figure that out by looking at the … Read more
Notice that space-compatible (radiation-hardened, aeronautics compliant) computing devices are very expensive (including to launch in space, since their weight exceeds kilograms), and that a single space mission costs perhaps hundred million € or US$. Losing the mission because of software or computer concerns has generally a prohibitive cost so is unacceptable and justifies costly development … Read more