Template Typename T

Template pointer parameter (passing a pointer to a function)</p> // template template parameter t has a parameter list, which // consists of one type template parameter with a default template<template<typename = float> typename t> struct a { void f(); You do, however, have to use class (and not typename) when declaring a template template parameter: Template struct check means a that template arguments are. Template struct derived_interface_type { typedef typename interface<derived, value> type;

You need one derived_interface_type for each instantiation of the derived template unfortunately, unless there's another trick i haven't learned yet.</p> The second one you actually show in your question, though you might not realize it: Template struct container { t t; Template it denotes a template which depends on a type t and a value t of that type.

// Class Template, With A Type Template Parameter With A Default Template Struct B {};

Template it denotes a template which depends on a type t and a value t of that type. Template pointer parameter (passing a pointer to a function)</p> Check* is a little bit more confusing.</p> The second one you actually show in your question, though you might not realize it:

Template< Typename T > Void Foo( T& X, Std::string Str, Int Count ) { // These Names Are Looked Up During The Second Phase // When Foo Is Instantiated And The Type T Is Known X.size();

The notation is a bit heavy since in most situations the type could be deduced from the value itself. Template < template < typename, typename > class container, typename type > This really sounds like a good idea though, if someone doesn't want to use type_traits. // pass 3 as argument.

Template Typename T> Class C { };

Like someone mentioned the main logic can be done in a different function, which accepts an extra flag to indicate the type, and this specialized declaration can just set the flag accordingly and directly pass on all the other arguments without touching anything. Template class foo { typedef typename param_t::baz sub_t; If solely considering this, there are two logical approaches: Template struct vector { unsigned char bytes[s];

Template Class T> Class C { };

Let's firstly cover the declaration of struct check; Template struct check means a that template arguments are. Typename and class are interchangeable in the declaration of a type template parameter. // template template parameter t has a parameter list, which // consists of one type template parameter with a default template<template<typename = float> typename t> struct a { void f();

Template pointer parameter (passing a pointer to a function)</p> You need one derived_interface_type for each instantiation of the derived template unfortunately, unless there's another trick i haven't learned yet.</p> Template< typename t > void foo( t& x, std::string str, int count ) { // these names are looked up during the second phase // when foo is instantiated and the type t is known x.size(); // pass type long as argument. You do, however, have to use class (and not typename) when declaring a template template parameter: