Template explicit instantiation of a method with a different body

Question

I have a template class with a constructor that takes a std::vector<T>. For every object except one I want it to make operation A. But for that one object, I want it to do some other stuff B.

Is there a possibility to create an explicit instantiation of only the constructor for a template class? I hope its described precisely enough.

Regards

Update: I have now implemented a test case:

//header

Container(const std::vector<T>& source)
{...}

//source code

template <> Container<int>::Container(const std::vector<int>& source)
{
    throw 42;
}

This example compiles but doesn't work. I export this into a dll and want to have it being called whenever I try to create an instance of the class with generic parameter int. But as it is now, it only calls the standard constructor used for every other object. Is there a change I have to make to the declaration?

Update: I succeeded! Just had to copy it to the header file.

Update: OK, now I have another problem. I am able to make a specialization for 'simple' types but not for templates. I tried it this way:

template<typename T>
Container<MyClass<T>>::Container(const std::vecror<MyClass<T>>& source)
{...}

I want to specialize it for every MyClass object, but MyClass itself shall be able to live on as a template.

It is not enough. Can you post at least pseudo-code what you want to do? — BЈовић, Jun 22 '11 at 07:38
*create an explicit instantiation*? you probably mean create an explicit **specialization**. You don't create instantiations, you can trigger instantiation either implicit or explicitly, but they are not *created*. — David Rodríguez - dribeas, Jun 22 '11 at 08:08

Oliver Charlesworth · Accepted Answer · 2011-06-22T07:58:05.223

5

Your question isn't clear. Perhaps you mean something like the following?

template <typename T>
class Foo
{
public:
    Foo() { std::cout << "standard" << std::endl; }
};

template <>
Foo<float>::Foo() { std::cout << "random" << std::endl; }  // Special case

...

Foo<int>    f1;  // Prints "standard"
Foo<float>  f2;  // Prints "random"

edited Jun 22 '11 at 07:58

answered Jun 22 '11 at 07:40

Oliver Charlesworth

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That's exactly what I meant. Thank you! PS: I'm using visual studio 2005. – Marcel Bonzelet Jun 22 '11 at 09:20

VGE · Answer 2 · 2011-06-22T07:49:46.300

0

Which compiler? I have had trouble with template instanciation with some old version of g++ compiler and with Solaris and HPUX c++ compiler. The template <> is used to specify explicit instanciation. It is a long time I do not have specialized only a method of a class.

Did you try ?

template <> TemplateClass<InstanciedType>::TemplateClass()  {
   ...
}

Where TemplateClass is the template class instanciation you want to override.

edited Jun 22 '11 at 07:49

answered Jun 22 '11 at 07:43

VGE

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`template<>` does the specialization, you might be referring to `template` for explicit instantiation purposes – Nick Jun 22 '11 at 07:45

score 0 · Answer 3 · edited May 23 '17 at 11:43

0

There is no way to do explicit constructor template instantiation -- this was covered here: C++ invoke explicit template constructor

edited May 23 '17 at 11:43

Community

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answered Jun 22 '11 at 07:50

Nick

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score 0 · Answer 4 · answered Jun 22 '11 at 07:58

0

If I understood your question correctly, you want to do this :

template< typename T >
struct A
{
  template< typename P >
  A( std::vector< P > &v )
  {
  }
};

answered Jun 22 '11 at 07:58

BЈовић

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Template explicit instantiation of a method with a different body

4 Answers4