[SOLVED] Question Regarding C++ Template Specialization.

BLXCryptor · 02-06-2011, 07:07 AM

Good Afternoon Guys,

I am having trouble understanding some concepts regarding C++ template specialization and I was hopping you guys can help me like you did last time.

I will go right into the code and let the code do the talking.

File foo.hpp

Code:

// Template Class
template < typename T1, typename T2> struct foo {
	static const char* str;
	static const char* getStr();
};

// str template initialisation.
template < typename T1, typename T2>
const char* foo<T1, T2>::str = "str<T1   , T2   >";

// str template specialisation.
template < >
const char* foo<float, float>::str = "str<float, float>";

// getStr template implementation.
template < typename T1, typename T2>
const char* foo<T1, T2>::getStr()
{ return "getStr<T1   , T2   >"; }

// getStr template specialisation.
template < >
const char* foo<float, float>::getStr()
{ return "getStr<float, float>"; }

File main.cpp

Code:

#include <iostream>
#include "foo.hpp"
using namespace std;

int main()
{
	cout   << "For <int  , int  > foo has " << foo<int  , int  >::str
		<< " and " << foo<int  , int  >::getStr() << endl;
	cout   << "For <float, float> foo has " << foo<float, float>::str
		<< " and " << foo<float, float>::getStr() << endl;
	cout   << "For <int  , float> foo has " << foo<int  , float>::str
		<< " and " << foo<int  , float>::getStr() << endl;
	cout   << "For <float, int  > foo has " << foo<float, int  >::str
		<< " and " << foo<float, int  >::getStr() << endl;
	return 0;
}

So far so good. All works fine and the output as expected is:

For <int , int > foo has str<T1 , T2 > and getStr<T1 , T2 >
For <float, float> foo has str<float, float> and getStr<float, float>
For <int , float> foo has str<T1 , T2 > and getStr<T1 , T2 >
For <float, int > foo has str<T1 , T2 > and getStr<T1 , T2 >

Now here is where I try to get funky and mess things up.
I want to do a specialization for just one of the two parameters and add the following code into foo.hpp

Code:

template < typename T1 >
const char* foo<T1, float>::getStr()
{ return "getStr<T1   , float>"; }

Which gives the error
error: invalid use of incomplete type ‘struct foo<T1, float>’

Which of course means that I am missing the following template specialization:

Code:

template < typename T1 > struct foo < T1, float>
{
	static const char* str;
	static const char* getStr();
};

However if I add this I have duplicated the definition of foo, which as you can see in both the main and the specialization is exactly the same; and now I also have to duplicate the source implementation for the new template specialization. For this example of course it is a simple initialization of str, however in a real program it could had been a lot more source.

I know that I can get thing behaving better if I create helper classes that all depend in one parameter, however I was wondering if I am missing something, or if there are other solutions to the above problem that I have not yet find, or if they have address this in C++0x.

Can you enlighten me?

Thanks a lot guys!

ta0kira · 02-06-2011, 10:44 PM

One option is to use a "traits"-type system, e.g. iterator_traits:

Code:

#include <iostream>


template <class Type1, class Type2>
struct specialized
{
	static const char name[];
};

template <class Type1, class Type2>
const char specialized <Type1, Type2> ::name[] = "<Type1, Type2>";


template <>
struct specialized <float, float>
{
	static const char name[];
};

const char specialized <float, float> ::name[] = "<float, float>";


template <class Type1>
struct specialized <Type1, float>
{
	static const char name[];
};

template <class Type1>
const char specialized <Type1, float> ::name[] = "<Type1, float>";


template <class Type1, class Type2>
struct some_class
{
	static const char *const name;

	const char *get_name() const
	{ return name; }
};

template <class Type1, class Type2>
const char *const some_class <Type1, Type2> ::name = specialized <Type1, Type2> ::name;


int main()
{
	some_class <int,   int>   int_int;
	some_class <int,   float> int_float;
	some_class <float, float> float_float;
	std::cout << "int_int: "     << int_int.get_name()     << "\n";
	std::cout << "int_float: "   << int_float.get_name()   << "\n";
	std::cout << "float_float: " << float_float.get_name() << "\n";
}

Something like that allows you to minimize explicit specialization if the only differences are typedefs and static data.

If you need to do something like that for function specialization, you can do that with static functions, although it can become a bit ugly if you need access to a lot of members in the functions:

Code:

#include <iostream>


template <class Type1, class Type2>
struct specialized
{
	static void show_values(const Type1 &vValue1, const Type2 &vValue2)
	{ std::cout << "Type1: " << vValue1 << ", Type2: " << vValue2 << "\n"; }
};


template <>
struct specialized <float, float>
{
	static void show_values(const float &vValue1, const float &vValue2)
	{ std::cout << "float: " << vValue1 << ", float: " << vValue2 << "\n"; }
};


template <class Type1>
struct specialized <Type1, float>
{
	static void show_values(const Type1 &vValue1, const float &vValue2)
	{ std::cout << "Type1: " << vValue1 << ", float: " << vValue2 << "\n"; }
};


template <class Type1, class Type2>
struct some_class
{
	some_class() : value1(), value2() {}

	void show_values() const
	{ specialized <Type1, Type2> ::show_values(value1, value2); }

	Type1 value1;
	Type2 value2;
};


int main()
{
	some_class <int,   int>   int_int;
	some_class <int,   float> int_float;
	some_class <float, float> float_float;
	int_int.show_values();
	int_float.show_values();
	float_float.show_values();
}

I think the sort of specialization you're talking about can't be done with a single class definition; therefore, you're better off finding a way to confine what's specialized to a class that only contains specialized members.
Kevin Barry

BLXCryptor · 02-10-2011, 12:23 PM

Thanks a lot! Your answer did help me!