In short, CRTP is when a class A has a base class which is a template specialization for the class A itself. E.g. template <class T> class X{…}; class A : public X<A> {…}; It is curiously recurring, isn’t it? 🙂 Now, what does this give you? This actually gives the X template the ability … Read more
<% %> Executes the ruby code within the brackets. <%= %> Prints something into erb file. <%== %> Equivalent to <%= raw %>. Prints something verbatim (i.e. w/o escaping) into erb file. (Taken from Ruby on Rails Guides.) <% -%> Avoids line break after expression. <%# %> Comments out code within brackets; not sent to … Read more
Short answer: in order to make x a dependent name, so that lookup is deferred until the template parameter is known. Long answer: when a compiler sees a template, it is supposed to perform certain checks immediately, without seeing the template parameter. Others are deferred until the parameter is known. It’s called two-phase compilation, and … Read more
Templated check for the existence of a class member function?
(See here also for my C++11 answer) In order to parse a C++ program, the compiler needs to know whether certain names are types or not. The following example demonstrates that: t * f; How should this be parsed? For many languages a compiler doesn’t need to know the meaning of a name in order … Read more
Caveat: It is not necessary to put the implementation in the header file, see the alternative solution at the end of this answer. Anyway, the reason your code is failing is that, when instantiating a template, the compiler creates a new class with the given template argument. For example: template<typename T> struct Foo { T … Read more
Well… Double_node<Type *> *ptr = stack.list.head(); is declaring a variable ptr which is of type pointer (thanks to the *) to Double_node<Type *> while Double_node<TType *> ptr = stack.list.head(); is declaring a variable ptr which is a Double_node<TType *>. The latter is not a pointer.
The template instantiated will be int* f(int* x, int y) { return x + y; } Therefore you’ll be doing pointer arithmetic, then dereferencing a pointer in which you did not allocate, which happens to be incremented 23 addresses past a. It is essentially as if you’d called *(new int(23) + 23);
there is no semicolon after: return Point <var> (x,y)
Let’s find out: <?php ob_start(); $a = 0; $time1 = microtime(true); for ($i = 0; $i < 100000; $i++) { echo “<html><body>$a</body></html>”; } $time2 = microtime(true); for ($i = 0; $i < 100000; $i++) { ?> <html><body><?php echo $a; ?></body></html> <?php } $time3 = microtime(true); ob_end_clean(); echo ‘Just echo: ‘ . ($time2 – $time1) . … Read more