模板专业化是否有一个微妙的技巧,以便我可以将一个专门化应用于基本POD(当我说基本POD我不特别想要结构POD(但我会采取那个)).
template<typename T>
struct DoStuff
{
void operator()() { std::cout << "Generic\n";}
};
template<>
struct DoStuff</*SOme Magic*/>
{
void operator()() { std::cout << "POD Type\n";}
};
或者我是否必须为每个内置类型编写特化？
template<typename T>
struct DoStuff
{
void operator()() { std::cout << "Generic\n";}
};
// Repeat the following template for each of
// unsigned long long,unsigned long,unsigned int,unsigned short,unsigned char
// long long,long,int,short,signed char
// long double,double,float,bool
// Did I forget anything?
//
// Is char covered by unsigned/signed char or do I need a specialization for that?
template<>
struct DoStuff<int>
{
void operator()() { std::cout << "POD Type\n";}
};
单元测试.
int main()
{
DoStuff<int> intStuff;
intStuff(); // Print POD Type
DoStuff<std::string> strStuff;
strStuff(); // Print Generic
}
解决方法
如果您真的只想要基本类型而不是用户定义的POD类型,那么以下内容应该有效：
#include <iostream>
#include <boost/type_traits/integral_constant.hpp>
#include <boost/type_traits/is_fundamental.hpp>
#include <boost/type_traits/is_same.hpp>
template<typename T>
struct non_void_fundamental : boost::integral_constant<
bool,boost::is_fundamental<T>::value && !boost::is_same<T,void>::value
>
{ };
template<typename T,bool Enable = non_void_fundamental<T>::value>
struct DoStuff
{
void operator ()() { std::cout << "Generic\n"; } const
};
template<>
struct DoStuff<T,true>
{
void operator ()() { std::cout << "POD Type\n"; } const
};
如果您还想要用户定义的POD类型,那么请使用boost :: is_pod<>而不是non_void_fundamental<> (如果您使用的是C 11并且为了优化目的,请使用std :: is_trivially_copyable<>).