Partial Template Specialization

[Agent Mulder]

Hi group,

I have a problem with partial template specialization. In the code
below I have a template struct Music with one method, play(),
and three kinds of music, Jazz, Funk and Bach. When I specialize
Music<Bach>, I expect that the original play() method is available
in the specialization, but it is not. How can I fix this?

-X

struct Jazz{};
struct Funk{};
struct Bach{};
template<struct A>struct Music{void play(){}};
struct Music<Bach>{}; //partial template specialization
int main()
{
Music<Jazz>().play(); //OK
Music<Funk>().play(); //OK
Music<Bach>().play(); //error, Music<Bach>().play not declared
return 0;
}

[Dave]

"Agent Mulder" <mb*******************@home.nl> wrote in message
news:bo**********@news4.tilbu1.nb.home.nl...

Hi group,

I have a problem with partial template specialization. In the code
below I have a template struct Music with one method, play(),
and three kinds of music, Jazz, Funk and Bach. When I specialize
Music<Bach>, I expect that the original play() method is available
in the specialization, but it is not. How can I fix this?

-X

struct Jazz{};
struct Funk{};
struct Bach{};
template<struct A>struct Music{void play(){}};
struct Music<Bach>{}; //partial template specialization
int main()
{
Music<Jazz>().play(); //OK
Music<Funk>().play(); //OK
Music<Bach>().play(); //error, Music<Bach>().play not declared
return 0;
}

I'm sure others will be able to give a more complete answer than I, but in
looking at your code, your specialization for Bach does indeed not have a
play() member function. In fact, it doesn't have any member functions or
data members - it's an empty class! I believe I'm correct in saying that
your partial specialization will have only what you explicitly put in it.
I'll rely on others to correct me if I'm wrong, but I think what I said is
correct...

Also, template<struct A> should be either template<typename A> or
template<class A> (these two are equivalent).

[Andrey Tarasevich]

Agent Mulder wrote:

...
I have a problem with partial template specialization. In the code
below I have a template struct Music with one method, play(),
and three kinds of music, Jazz, Funk and Bach. When I specialize
Music<Bach>, I expect that the original play() method is available
in the specialization, but it is not. How can I fix this?
...
struct Jazz{};
struct Funk{};
struct Bach{};
template<struct A>struct Music{void play(){}};
struct Music<Bach>{}; //partial template specialization
This is not partial specialization. This is explicit specialization. And
the correct syntax is a s follows

template<> struct Music<Bach> {};
int main()
{
Music<Jazz>().play(); //OK
Music<Funk>().play(); //OK
Music<Bach>().play(); //error, Music<Bach>().play not declared
return 0;
}
...

Explicit (or partial) specialization of a template is an independently
defined template. You defined it as a template class with no explicit
members. That's what you get when you instantiate the specialized
version - a class with no explicit members. It doesn't have method
'play()', hence the error.

If you want to have an explicit specialization of the entire
'Music<Bach>' and still have method 'play' in it, you have no other
choice but to declare and define this method in 'Music<Bach>' explicitly.

template<> struct Music<Bach> { void play( /* whatever */ ); };

On the other hand, if you just want to explicitly specialize the method
'play' for each kind of music, there is no need to explicitly specialize
the entire template 'Music'

struct Jazz {};
struct Funk {};
struct Bach {};

template<struct A> struct Music { void play(){} };

// Explicit specializations for method 'play()'
template<> void Music<Jazz>::play() { /* whatever 1 */ }
template<> void Music<Funk>::play() { /* whatever 2 */ }
template<> void Music<Bach>::play() { /* whatever 3 */ }

--
Best regards,
Andrey Tarasevich

[Andrey Tarasevich]

Agent Mulder wrote:

...
template<struct A>struct Music{void play(){}};
...

BTW, something I haven't noticed right away: you cannot use keyword
'struct' to declare template parameters of a template. The only keywords
that can be used for this purpose are 'typename' and 'class'.

By using keyword 'struct' you are actually making an attempt to declare
a non-type parameter of type 'struct A'. This is illegal. Firstly,
non-type template parameters of class type are not allowed in C++. And
secondly, you don't even have a struct named 'A' in your program.

--
Best regards,
Andrey Tarasevich

[Cy Edmunds]

"Agent Mulder" <mb*******************@home.nl> wrote in message
news:bo**********@news4.tilbu1.nb.home.nl...

Hi group,

I have a problem with partial template specialization. In the code
below I have a template struct Music with one method, play(),
and three kinds of music, Jazz, Funk and Bach. When I specialize
Music<Bach>, I expect that the original play() method is available
in the specialization, but it is not. How can I fix this?

-X

struct Jazz{};
struct Funk{};
struct Bach{};
template<struct A>struct Music{void play(){}};
struct Music<Bach>{}; //partial template specialization
int main()
{
Music<Jazz>().play(); //OK
Music<Funk>().play(); //OK
Music<Bach>().play(); //error, Music<Bach>().play not declared
return 0;
}

Maybe they use templates this way on the X-Files but... hehe

I would suggest this design instead:

class Music
{
public:
virtual void play() = 0;
virtual ~Music() {}
};

class Jazz : public Music
{
public:
virtual void play();
};

class Funk: public Music
{
public:
virtual void play();
};

etc.

--
Cy
http://home.rochester.rr.com/cyhome/

[Agent Mulder]

Thank you for your insightful responses. I fixed the design
by letting the template base and the template specialization
both inherit from the same class CD. This class CD has a virtual
play() method that gets called through a reference.

Still I am not satisfied. First, I need to forward declare struct
Bach. Second, both templates need to inherit class CD. What's the
use of template specialization if you lose everything that was
specified in the base template?

#include<iostream>
struct Bach; //forward declaration
struct CD{virtual void play(){cout<<"\nFeep-Honk-Feep-Honk";}};
template<class A>struct Music:CD{};
struct Music<Bach>:CD{void play(){cout<<"\nHonk-Honk-Feep-Feep";}};
struct Jazz{};
struct Funk{};
struct Bach{};
Music<Jazz>jazz;
Music<Funk>funk;
Music<Bach>bach;
int main()
{
CD&jazz=::jazz;
CD&funk=::funk;
CD&bach=::bach;
jazz.play();
funk.play();
bach.play();
return 0;
}

-------------------
output:
Feep-Honk-Feep-Honk
Feep-Honk-Feep-Honk
Honk-Honk-Feep-Feep

-X

[Cy Edmunds]

"Agent Mulder" <mb*******************@home.nl> wrote in message
news:bo**********@news1.tilbu1.nb.home.nl...

Thank you for your insightful responses. I fixed the design
by letting the template base and the template specialization
both inherit from the same class CD. This class CD has a virtual
play() method that gets called through a reference.

Still I am not satisfied. First, I need to forward declare struct
Bach. Second, both templates need to inherit class CD. What's the
use of template specialization if you lose everything that was
specified in the base template?

What indeed? You don't need any templates to do what you are trying to do.
They just confuse the issue.

--
Cy
http://home.rochester.rr.com/cyhome/

[Agent Mulder]

"Cy Edmunds" <ce******@spamless.rochester.rr.com> wrote in message news:a0*******************@twister.nyroc.rr.com...

"Agent Mulder" <mb*******************@home.nl> wrote in message
news:bo**********@news1.tilbu1.nb.home.nl...

Thank you for your insightful responses. I fixed the design
by letting the template base and the template specialization
both inherit from the same class CD. This class CD has a virtual
play() method that gets called through a reference.

Still I am not satisfied. First, I need to forward declare struct
Bach. Second, both templates need to inherit class CD. What's the
use of template specialization if you lose everything that was
specified in the base template?

What indeed? You don't need any templates to do what you are trying to do.
They just confuse the issue.

Templates I try to understand, not the inner working of Music<Bach>::play();
although you do seem to think so. Refer to Stroustrup 13.6 Derivation and
Templates for examples of how to organize your code around templates.

-X

I try to understand templates I try to understand. Not to implement Music<Jazz>::play(),
although you do seem to think so.

[Jerry Coffin]

In article <bo**********@news4.tilbu1.nb.home.nl>,
mb*******************@home.nl says...

Hi group,

I have a problem with partial template specialization.
First and foremost that what you think is partial specialization is
really explicit specialization?

Partial specialization looks something like this:

template <class A, class B>
class X { ... };

template <class A>
class X<int, A> { ... };

The first is unspecialized -- either parameter can potentially be bound
to any type. The second is a partial specialization. We're specifying
what code will be produced when the first parameter happens to be int,
but the second parameter can still vary. The bottom line: to be partial
specialization, you have to have at least one each of specified
parameters and unspecified parameters. If you're specifying the types
of all parameters, you have an explicit specialization instead.
In the code
below I have a template struct Music with one method, play(),
and three kinds of music, Jazz, Funk and Bach. When I specialize
Music<Bach>, I expect that the original play() method is available
in the specialization, but it is not. How can I fix this?

No -- specialization is not inheritance. Each specialization (partial
or explicit) is independent of the primary template, and has to define
its own members. The members of a specialization may be completely
different from the members of the primary template.

--
Later,
Jerry.

The universe is a figment of its own imagination.