仅适用于基本 POD 的模板专业化

Question

Is there a subtle trick for template specialization so that I can apply one specialization to basic POD (when I say basic POD I don't particularly want struct POD (but I will take that)).

template<typename T>
struct DoStuff
{
    void operator()() { std::cout << "Generic\n";}
};
template<>
struct DoStuff</*SOme Magic*/>
{
    void operator()() { std::cout << "POD Type\n";}
};

Or do I have to write specializations for each of the built in types?

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";}
};

Unit Test.

int main()
{
    DoStuff<int>           intStuff;
    intStuff();            // Print POD Type


    DoStuff<std::string>   strStuff;
    strStuff();            // Print Generic
}

Answer 1

If you really want only fundamental types and not user-defined POD types then the following should work:

#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
};

If you also want user-defined POD types, then use boost::is_pod<> instead of non_void_fundamental<> (and if you're using C++11 and doing this for optimization purposes, use std::is_trivially_copyable<> instead).

Answer 2

In C++11, many traits have been added to the standard library, and most seem particularly aimed toward interesting specializations (and notably bitwise manipulations).

The top-level trait you could be interested in is std::is_trivial , however there are many others:

• std::is_trivially_default_constructible
• std::is_trivially_copy_constructible
• std::is_trivially_move_constructible
• std::is_trivially_copyable (can be copied via memcpy )

In general, the Standard has tried to get as finer grained traits as possible so you need not rely on such broad assumptions as is_pod but instead fine-tune your constraints to match what your methods really need.

Answer 3

Boost has boost::is_pod . Is that what you're looking for?

(I've never used it, so I won't embarrass myself by trying to formulate the precise code that you require for your example.)