我正在尝试为两种不同类型的类专门化成员函数模板,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>();
}
而 GCC 4.8.2 编译代码,VS .NET 2008 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)'