Specialized template method with derived classes as argument

Question

With a template method, that is specialized for a specific class A, is there a way to call the specialized code for the classes deriving from A ?

The minimal example

#include <iostream<  

class A
{};

class B : public A
{};

class Templator
{
public :
    template <class T>
    void dance(T *argument);
};

template <class T>
void Templator::dance(T *argument)
{
    std::cout << "General implementation" << std::endl;
}

template<>
void Templator::dance<A> (A* )
{
    std::cout << "Specialized implementation" << std::endl;
}

int main()
{
    A a;
    B b;

    Templator tt;
    tt.dance(&a);
    tt.dance(&b);

    return 0;
}

Corresponding output

Specialized implementation
General implementation

In other words, I would like the two outputs to be :

Specialized implementation
Specialized implementation

http://stackoverflow.com/questions/4984502/force-template-class-to-be-derived-from-baseclassa — Lol4t0, Jan 12 '12 at 15:34
You can have a look at this question I asked a while ago about this issue: http://stackoverflow.com/q/7928871/20984 — Luc Touraille, Jan 12 '12 at 17:12

score 3 · Accepted Answer · answered Jan 12 '12 at 15:41

3

In C++11 you can use std::enable_if and std::is_base_and_derived:

class Templator{
public :
    template <class T>
    typename std::enable_if<!std::is_base_and_derived<A, T>::value, void>::type 
    dance(T *argument) { 
       std::cout << "General implementation" << std::endl;
    }
    template <class T>
    typename std::enable_if<std::is_base_and_derived<A, T>::value, void>::type 
    dance(T *argument) { 
       std::cout << "Specialized implementation" << std::endl;
    }
};

You can do the same thing in C++03 using either boost::enable_if_c and boost::is_base_of or std::tr1::is_base_of and writing enable_if yourself (see here)

answered Jan 12 '12 at 15:41

Grizzly

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Thank you, I accepted your answer because it was illustrating the use of enable_if and is_base_of (that I obviously was not aware of the existence). And it is solving my specific problem, that's definitely a plus ; ) – Ad N Jan 12 '12 at 17:02
I was about to advise you to use `disable_if
` instead of `enable_if<!P>`, before realizing it was not included in C++11 (like the original `boost::enable_if` taking a metafunction instead of a boolean)... However, they included the default parameter for the second parameter, so you can omit `void` if you want to avoid some clutter.
– Luc Touraille Jan 12 '12 at 17:23
@LucTouraille: Personally I'd prefer explicitely stating the `void`, since even if it is a default parameter, I think explicitely mentioning the return type is a good idea to make the code clearer. – Grizzly Jan 12 '12 at 17:28
Sure, that's a tenable point of view :). I just made this remark for the OP's information. – Luc Touraille Jan 12 '12 at 17:32

score 1 · Answer 2 · answered Jan 12 '12 at 15:20

1

You just need to cast &b to a A*.

If you change this code

tt.dance(&a);
tt.dance(&b);

To this code

tt.dance(&a);
tt.dance((A*)&b);

The output becomes

Specialized implementation
Specialized implementation

answered Jan 12 '12 at 15:20

Jack Edmonds

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Yes, it works (like calling `tt.dance(&b)`), but I am sorry I oversimplified my problem. I would still need T to be resolved to B, because my real method implementation call different code in case T is an A or a B... and now that I phrases it, I understand that what I am asking is not possible because I want T to be a B and a A. Those Q&A websites are also good for self teaching ^^ – Ad N Jan 12 '12 at 15:25

score 1 · Answer 3 · answered Jan 12 '12 at 15:22

1

This should be possible using a boost::enable_if in conjunction with boost::is_base_of.

answered Jan 12 '12 at 15:22

Björn Pollex

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Yes, it diverges from pure "specialization" to use SFINAE instead... but this is a way to make it work (cf. Lol4t0 pointer under the question) – Ad N Jan 12 '12 at 15:53

Specialized template method with derived classes as argument

The minimal example

Corresponding output

3 Answers3