模板函数调用

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我在看boost的ASIO源代码的时候，发现一个没有见过的模板调用用法.google没有说明，baidu也没有说明，只好自己试一下。下面是ASIO调用的代码：在文件service_registry.hpp中。

1   // Check whether a service object of the specified type already exists.
2   template <typename Service>
3   bool has_service() const
4   {
5     boost::asio::detail::mutex::scoped_lock lock(mutex_);
6
7     boost::asio::io_service::service* service = first_service_;
8     while (service)
9     {
10       if (service_id_matches(*service, Service::id))
11         return true;
12       service = service->next_;
13     }
14
15     return false;
16   }

调用的地方,红色标出的地方

template <typename Service>
bool has_service(io_service& ios)
{
  // Check that Service meets the necessary type requirements.
  (void)static_cast<io_service::service*>(static_cast<Service*>(0));
  (void)static_cast<const io_service::id*>(&Service::id);

  return ios.service_registry_->template has_service<Service>();
}

下面是测试例子，在VS2005中编译通过。原来，函数使用的时候，可以完全指定模板函数定义。

1 class VV
2 {
3 public:
4     static int getI()
5     {
6         return 100;
7     }
8 };
9 class V
10 {
11 public:
12     template<class K>
13     void PrintK()
14     {
15         cout<<K::getI()<<endl;
16     }
17 };
18 void m()
19 {
20     V v;
21     v.template PrintK<VV>();
22 }

ANSI C++ '03
14.2 Names of template specializations
4 When the name of a member template specialization appears after . or -> in a postfix-expression, or after
nested-name-specifier in a qualified-id, and the postfix-expression or qualified-id explicitly depends on a
template-parameter (14.6.2), the member template name must be prefixed by the keyword template.
Otherwise the name is assumed to name a non-template. [Example:
class X {
public:
template<size_t> X* alloc();
template<size_t> static X* adjust();
};
template<class T> void f(T* p)
{
T* p1 = p->alloc<200>();
// ill-formed: < means less than
T* p2 = p->template alloc<200>();
// OK: < starts template argument list
T::adjust<100>();
// ill-formed: < means less than
T::template adjust<100>();
// OK: < starts template argument list
}

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