静态成员函数模板专用化;如何？ [英] template specialization for static member functions; howto?

问题描述

我试图实现与手柄无效不同使用模板特殊化模板函数。

I am trying to implement a template function with handles void differently using template specialization.

下面的代码给我一个非命名空间范围明确的专业化的gcc：

The following code gives me an "Explicit specialization in non-namespace scope" in gcc:

template <typename T>
static T safeGuiCall(boost::function<T ()> _f)
{
	if (_f.empty())
		throw GuiException("Function pointer empty");
	{
		ThreadGuard g;
		T ret = _f();
		return ret;
	}
}

// template specialization for functions wit no return value
template <>
static void safeGuiCall<void>(boost::function<void ()> _f)
{
	if (_f.empty())
		throw GuiException("Function pointer empty");
	{
		ThreadGuard g;
		_f();
	}
}

我已经尝试将它移出类类不是模板）和进入命名空间，但然后我得到错误显式专业化不能有一个存储类。我已经阅读了很多关于这一点的讨论，但是人们似乎不同意如何专门化功能模板。 ？任何想法

I have tried moving it out of the class (the class is not templated) and into the namespace but then I get the error "Explicit specialization cannot have a storage class". I have read many discussions about this, but people don't seem to agree how to specialize function templates. Any ideas?

推荐答案

当你专门一个模板方法，你必须与类括号外面这样做的：

When you specialize a templated method, you must do so outside of the class brackets:

template <typename X> struct Test {}; // to simulate type dependency

struct X // class declaration: only generic
{
   template <typename T>
   static void f( Test<T> );
};

// template definition:
template <typename T>
void X::f( Test<T> ) {
   std::cout << "generic" << std::endl;
}
template <>
inline void X::f<void>( Test<void> ) {
   std::cout << "specific" << std::endl;
}

int main()
{
   Test<int> ti;
   Test<void> tv;
   X::f( ti ); // prints 'generic'
   X::f( tv ); // prints 'specific'
}

必须删除static关键字。 ，在课堂外static关键字拥有你可能想要什么不同的特定含义。

When you take it outside of the class, you must remove the 'static' keyword. Static keyword outside of the class has a specific meaning different from what you probably want.

template <typename X> struct Test {}; // to simulate type dependency

template <typename T>
void f( Test<T> ) {
   std::cout << "generic" << std::endl;
}
template <>
void f<void>( Test<void> ) {
   std::cout << "specific" << std::endl;
}

int main()
{
   Test<int> ti;
   Test<void> tv;
   f( ti ); // prints 'generic'
   f( tv ); // prints 'specific'
}

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