[英]Unable to specialize a member function template with boost::enable_if in VS .NET 2008
I'm trying to specialize a member function template for two different types of classes as follows: 我试图专门针对两种不同类型的类的成员函数模板,如下所示:
#include <iostream>
#include <boost/utility/enable_if.hpp>
struct Wibble
{
static const bool CAN_WIBBLE = true;
};
struct Wobble
{
static const bool CAN_WIBBLE = false;
};
struct Foo
{
//template<typename T> // Why isn't this declaration sufficient?
//void doStuff();
template<typename T>
typename boost::enable_if_c<T::CAN_WIBBLE,void>::type
doStuff();
template<typename T>
typename boost::enable_if_c<!T::CAN_WIBBLE,void>::type
doStuff();
};
template<typename T>
typename boost::enable_if_c<T::CAN_WIBBLE,void>::type
Foo::doStuff()
{
std::cout << "wibble ..." << std::endl;
}
template<typename T>
typename boost::enable_if_c<!T::CAN_WIBBLE,void>::type
Foo::doStuff()
{
std::cout << "I can't wibble ..." << std::endl;
}
int main()
{
Foo f;
f.doStuff<Wibble>();
f.doStuff<Wobble>();
}
Whereas GCC 4.8.2 compiles the code, VS .NET 2008 spits out the error message: GCC 4.8.2会编译代码,而VS .NET 2008会吐出错误消息:
error C2244: 'Foo::doStuff' : unable to match function definition to an existing declaration
definition
'boost::enable_if_c<!T::CAN_WIBBLE,void>::type Foo::doStuff(void)'
existing declarations
'boost::enable_if_c<!T::CAN_WIBBLE,void>::type Foo::doStuff(void)'
'boost::enable_if_c<T::CAN_WIBBLE,void>::type Foo::doStuff(void)'
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struct Foo
{
template<bool wibble>
void _doStuff();
public:
template<typename T>
void doStuff()
{
_doStuff<T::CAN_WIBBLE>();
}
};
template<>
void Foo::_doStuff<true>() { std::cout << "wibble ..." << std::endl; }
template<>
void Foo::_doStuff<false>() { std::cout << "I can't wibble ..." << std::endl; }
You can't partially specialize (member) function templates. 您不能部分专门化(成员)功能模板。 End of story.
故事结局。
Even if you could, you should have had a SFINAE-friendly primary template. 即使可以,您也应该拥有一个对SFINAE友好的主模板。 In pseudo code:
用伪代码:
template<typename T, typename Enable> void doStuff();
template<typename T> void doStuff<T, typename boost::enable_if_c<T::CAN_WIBBLE,void>::type>()
{ std::cout << "wibble ..." << std::endl; }
template<typename T> void doStuff<T, typename boost::enable_if_c<!T::CAN_WIBBLE,void>::type>()
{ std::cout << "I can't wibble ..." << std::endl; }
You could still use this technique if you are ready class templates (as functors or just types defining non-template methods...). 如果您已经准备好类模板(例如仿函数或仅定义非模板方法的类型...),则仍可以使用此技术。
As a rule of thumb, for function templates , overload resolution provides static polymorphism that removes the need for partial specialization. 根据经验,对于功能模板 ,重载解析提供了静态多态性,从而消除了部分专业化的需求。 See
看到
Both by Herb Sutter 两者均来自Herb Sutter
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