When is the "typename" keyword necessary?
Possible Duplicate:
Officially, what is typename for?
Where and why do I have to put the template and typename keywords?
consider the code below:
template<class K>
class C {
struct P {};
vector<P> vec;
void f();
};
template<class K> void C<K>::f() {
typename vector<P>::iterator p = vec.begin();
}
Why is the "typename" keyword necessary in this example? Are there any other cases where "typename" must be specified?
Short answer: Whenever referring to a nested name that is a dependent name, ie nested inside a template instance with unknown parameter.
Long answer: There are three tiers of entities in C++: values, types, and templates. All of those can have names, and the name alone doesn't tell you which tier of entity it is. Rather, the information about the nature of a name's entity must be inferred from the context.
Whenever this inference is impossible, you have to specify it:
template <typename> struct Magic; // defined somewhere else
template <typename T> struct A
{
static const int value = Magic<T>::gnarl; // assumed "value"
typedef typename Magic<T>::brugh my_type; // decreed "type"
// ^^^^^^^^
void foo() {
Magic<T>::template kwpq<T>(1, 'a', .5); // decreed "template"
// ^^^^^^^^
}
};
Here the names Magic<T>::gnarl , Magic<T>::brugh and Magic<T>::kwpq had to be expliciated, because it is impossible to tell: Since Magic is a template, the very nature of the type Magic<T> depends on T -- there may be specializations which are entirely different from the primary template, for example.
What makes Magic<T>::gnarl a dependent name is the fact that we're inside a template definition, where T is unknown. Had we used Magic<int> , this would be different, since the compiler knows (you promise!) the full definition of Magic<int> .
(If you want to test this yourself, here's a sample definition of Magic that you can use. Pardon the use of constexpr in the specializaation for brevity; if you have an old compiler, feel free to change the static member constant declaration to the old-style pre-C++11 form.)
template <typename T> struct Magic
{
static const T gnarl;
typedef T & brugh;
template <typename S> static void kwpq(int, char, double) { T x; }
};
template <> struct Magic<signed char>
{
// note that `gnarl` is absent
static constexpr long double brugh = 0.25; // `brugh` is now a value
template <typename S> static int kwpq(int a, int b) { return a + b; }
};
Usage:
int main()
{
A<int> a;
a.foo();
return Magic<signed char>::kwpq<float>(2, 3); // no disambiguation here!
}
The typename keyword, is needed because iterator is a dependent type on P . The compiler can't guess if iterator refers to a value or a type, so it assume its a value unless you yell typename . It's needed whenever there is a type dependent on a template argument, in a context that either types or values would be valid. For instance, as base classes typename is not needed since a base class must be a type.
On the same subject, there is a template keyword used to let the compiler know that some dependent name is a template function instead of a value.
The typename keyword is needed whenever a type name depends on a template parameter, (so the compiler can 'know' the semantics of an identifier (type or value) without having a full symbol table at the first pass).
Not in the same meaning, and a bit less common, the lone typename keyword can also be useful when using generic template parameters: http://ideone.com/amImX
#include <string>
#include <list>
#include <vector>
template <template <typename, typename> class Container,
template <typename> class Alloc = std::allocator>
struct ContainerTests
{
typedef Container<int, Alloc<int> > IntContainer;
typedef Container<std::string, Alloc<int> > StringContainer;
//
void DoTests()
{
IntContainer ints;
StringContainer strings;
// ... etc
}
};
int main()
{
ContainerTests<std::vector> t1;
ContainerTests<std::list> t2;
t1.DoTests();
t2.DoTests();
}
上一篇: 什么是Linux的本地GUI API？
下一篇: 何时需要“typename”关键字？