asked on

Template Specialization of Nested Class

I understand that template specializations are only valid in the namespace scope. So, in order to specialize a nested class, you need to define the specialization outside the class declaration.

Here, I try to specialize a nested class inside a templated class:

template <class T>
class A
{
template <bool B>
struct Test;
};

template <class T>
template<>
struct A<T>::Test<true> { };

However, the above code doesn't compile. Is there something wrong with my syntax?

SOLUTION

SteveH_UK

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

ASKER CERTIFIED SOLUTION

jkr

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

ikework

either you do the specialization inside of the class:

template <class T>
class A
{
public:
template <class B>
struct Test
{
B val;
};

template <>
struct Test<bool>
{
bool val;
};
};

or you need to specialize both template-parameters:

template <class T>
class A
{
public:
template <class B>
struct Test
{
B val;
};

};

template<>
template<>
struct A<int>::Test<bool>
{
bool val;
};

ike

chsalvia

ASKER

>> either you do the specialization inside of the class:

I don't think you can do this. Template specialization only seems to work in the namespace scope.

SteveH_UK

In my view, none of this would be good code:

1) it is not readable
2) it really doesn't serve any purpose, certainly not one that could be done differently
3) it may use some Microsoft specific extensions

ikework

>> "'specialization' : an explicit specialization of a template member must be a member of an explicit specialization"

mh .. the following works with VC8:

template <class T>
class A
{
public:
template <class B>
struct Test
{
Test() { printf( "no specialization\n" ); }
B val;
};

template <>
struct Test<bool>
{
Test() { printf( "specialization called\n" ); }
bool val;
};
};

int main
{
class A<int>::Test<bool> C;
class A<int>::Test<int> D;

return 0;
}

output is:

specialization called
no specialization

so it seems to work ..

ike

chsalvia

ASKER

It doesn't work on GCC 4. Maybe it's a Microsoft extension

ikework

>> >> either you do the specialization inside of the class:

>>I don't think you can do this. Template specialization only seems to work in the namespace scope.

run the code from my last post .. ;)

ikework

>> It doesn't work on GCC 4. Maybe it's a Microsoft extension

can you post the error-message?

evilrix

>> the following works with VC8
It's wrong, gcc4 is right! The standard states the specialization should only take place at namespace level -- VC8 (like in many other areas) ignores this!

>> can you post the error-message?
error: explicit specialization in non-namespace scope 'class A<T>'

The compiler will not allow the template specialization of Test because the enclosing class of class A is not explicitly specialized. Basically, as jkr has already stated, you need to specialize the outer before you can specialize the inner.

error C3212: 'A<T>::Test<true>' : an explicit specialization of a template member must be a member of an explicit specialization

evilrix

You can most likely achieve what you want by partial specialization of A, thus: -

#include <iostream>
 
template <class T, bool B>
struct A
{
	struct Test;
};
 
template <class T>
struct A<T, true>
{
	struct Test
	{
		static bool foo() { return true; }
	};
};
 
template <class T>
struct A<T, false>
{
	struct Test
	{
		static bool foo() { return false; }
	};
};
 
int main ()
{
	std::cout << "true = " << A<int, true>::Test::foo() << std::endl;
	std::cout << "false = "  << A<int, false>::Test::foo() << std::endl;
}

Open in new window

SOLUTION

evilrix

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial