Re: knowing the instance of a template when compiling the destructor of a class

From:
=?UTF-8?B?RXJpayBXaWtzdHLDtm0=?= <Erik-wikstrom@telia.com>
Newsgroups:
comp.lang.c++
Date:
Thu, 01 Nov 2007 10:14:11 GMT
Message-ID:
<TFhWi.12654$ZA.8150@newsb.telia.net>
On 2007-11-01 10:50, nick4ng@googlemail.com wrote:

If I have a templated constructor, there is a way to know, at compile
time, the instance when compiling the destructor?

To be clearer my problem is that I have a templated constructor that
is empty for particular arguments. I need to have an empty destructor
in this case as well.

Is it possible?


You mean something like this:

#include <iostream>

template<typename T>
class Foo
{
  T* t;
public:
  Foo();
  ~Foo();
};

template<typename T>
Foo<T>::Foo()
{
  t = new T(); // Should use initialisation list
}

template<typename T>
Foo<T>::~Foo()
{
  delete t;
}

template<>
Foo<int>::Foo()
{
}

template<>
Foo<int>::~Foo()
{
}

int main()
{
  Foo<double> d;
  Foo<int> i;
}

When instantiating Foo with an int the special constructor (and later
destructor) will be used. Notice however that I think it is a bad idea
to specialise the destructor if it can be avoided. It is better to just
set t to 0 in Foo<int>'s constructor and not specialise the destructor.

--
Erik Wikstr??m

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