Pure virtual destructor in template class

[Tonni Tielens]

I'm trying to create a pure virtual class describing an interface.
Normally, when I do this I make the destructor pure virtual so that,
even if there are no members in the class, it cannot be instantiated.

The difference now is that I'm making a generic interface with
template arguments. Template classes should be defined in the header
file, but it is not allowed for a destructor's definition to be in the
class definition if the destructor is pure virtual. Atleast not with
GCC -pedantic and I understand this is correct behavior. I'm unsure on
how to solve this. I know I don't really have to put a pure virtual
destructor in the class, but I think it's good practice so if it's
possible I would like to stick to this.

My code looks like the following:


template <typename TypeA, typename TypeB>
struct MyInterface
{
virtual ~MyInterface() = 0;

virtual void Foo(TypeA a, TypeB b) = 0;
};


I already found that I can resolve the compilation errors, by adding


template <typename TypeA, typename TypeB>
MyInterface<Typ eA, TypeB>::~MyInte rface() {}


in the same file after the class definition, but I'm not sure if this
is the common way to do this. Is there a correct way to do this or
should I leave the pure virtual destructor out?

[Victor Bazarov]

Re: Pure virtual destructor in template class

Tonni Tielens wrote:

I'm trying to create a pure virtual class describing an interface.
Normally, when I do this I make the destructor pure virtual so that,
even if there are no members in the class, it cannot be instantiated.

Why would you have an interface with no other members? Wouldn't it be
pretty much useless as an interface?

The difference now is that I'm making a generic interface with
template arguments. Template classes should be defined in the header
file, but it is not allowed for a destructor's definition to be in the
class definition if the destructor is pure virtual. Atleast not with
GCC -pedantic and I understand this is correct behavior. I'm unsure on
how to solve this. I know I don't really have to put a pure virtual
destructor in the class, but I think it's good practice so if it's
possible I would like to stick to this.
>
My code looks like the following:
>
>
template <typename TypeA, typename TypeB>
struct MyInterface
{
virtual ~MyInterface() = 0;
>
virtual void Foo(TypeA a, TypeB b) = 0;
};
>
>
I already found that I can resolve the compilation errors, by adding
>
>
template <typename TypeA, typename TypeB>
MyInterface<Typ eA, TypeB>::~MyInte rface() {}
>
>
in the same file after the class definition, but I'm not sure if this
is the common way to do this. Is there a correct way to do this or
should I leave the pure virtual destructor out?

If your destructor doesn't do anything, you can give it an empty body
right in the class definition. Since you have other virtual functions
in your interface, they will be pure and the destructor doesn't have to
be. But if you just *want* your destructor pure, your solution is just
what the doctor ordered. Idiomatic.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Pete Becker]

Re: Pure virtual destructor in template class

On 2008-11-18 11:33:49 -0500, Tonni Tielens <tonnitielens@g mail.comsaid:

I'm trying to create a pure virtual class describing an interface.

There's no such thing as a pure virtual class. The term you're looking
for is abstract class.

Normally, when I do this I make the destructor pure virtual so that,
even if there are no members in the class, it cannot be instantiated.
>
The difference now is that I'm making a generic interface with
template arguments. Template classes should be defined in the header
file, but it is not allowed for a destructor's definition to be in the
class definition if the destructor is pure virtual. Atleast not with
GCC -pedantic and I understand this is correct behavior. I'm unsure on
how to solve this. I know I don't really have to put a pure virtual
destructor in the class, but I think it's good practice so if it's
possible I would like to stick to this.
>
My code looks like the following:
>
>
template <typename TypeA, typename TypeB>
struct MyInterface
{
virtual ~MyInterface() = 0;
>
virtual void Foo(TypeA a, TypeB b) = 0;
};
>
>
I already found that I can resolve the compilation errors, by adding
>
>
template <typename TypeA, typename TypeB>
MyInterface<Typ eA, TypeB>::~MyInte rface() {}
>
>
in the same file after the class definition, but I'm not sure if this
is the common way to do this.

It is.

Is there a correct way to do this or
should I leave the pure virtual destructor out?

You can't leave the destructor out. Try it.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Pete Becker]

Re: Pure virtual destructor in template class

On 2008-11-18 11:52:08 -0500, Victor Bazarov <v.Abazarov@com Acast.netsaid:

Tonni Tielens wrote:

>I'm trying to create a pure virtual class describing an interface.
>Normally, when I do this I make the destructor pure virtual so that,
>even if there are no members in the class, it cannot be instantiated.

>
Why would you have an interface with no other members? Wouldn't it be
pretty much useless as an interface?

It's a Java thing. For example, if a class that implements
java.util.List (which is, roughly, a linked list) also implements
java.util.Rando mAccess, it announces that it supports constant-time
random access to elements, so a loop like this:

for (int i = 0, n = list.size(); i <n; i++)
list.get(i);

will supposedly be faster than a loop like this:

for (Iterator i = list.iterator() ; i.hasNext(); )
i.next();

You'd use a runtime type check to decide which way to go.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Tonni Tielens]

Re: Pure virtual destructor in template class

On Nov 18, 5:52 pm, Victor Bazarov <v.Abaza...@com Acast.netwrote:

Why would you have an interface with no other members?  Wouldn't it be
pretty much useless as an interface?

It's just a way (and I believe a common way) of telling a class is
abstract without having the need to have any abstract members.
Consider Java or C# where you can define an interface without having
any methods. Completely useless, but perfectly legal. Normally, since
it is almost no effort, I make the destructor pure virtual, but since
it is more effort here and I'm going to add pure virtual methods
anyway, I will leave it out and let the compiler generate a default
one.

Thanks for the replies.

[Victor Bazarov]

Re: Pure virtual destructor in template class

Tonni Tielens wrote:

On Nov 18, 5:52 pm, Victor Bazarov <v.Abaza...@com Acast.netwrote:

>Why would you have an interface with no other members? Wouldn't it be
>pretty much useless as an interface?

>
It's just a way (and I believe a common way) of telling a class is
abstract without having the need to have any abstract members.
Consider Java or C# where you can define an interface without having
any methods. Completely useless, but perfectly legal. Normally, since
it is almost no effort, I make the destructor pure virtual, but since
it is more effort here and I'm going to add pure virtual methods
anyway, I will leave it out and let the compiler generate a default
one.

No, don't let the compiler do it because in that case it wouldn't be
virtual. You *do* need the destructor to be virtual, trust me.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Tonni Tielens]

Re: Pure virtual destructor in template class

On Nov 18, 7:42 pm, Victor Bazarov <v.Abaza...@com Acast.netwrote:

No, don't let the compiler do it because in that case it wouldn't be
virtual.  You *do* need the destructor to be virtual, trust me.

Ah, you're completely right. I wasn't thinking while posting that. :)

[James Kanze]

Re: Pure virtual destructor in template class

On Nov 18, 6:06 pm, Pete Becker <p...@versatile coding.comwrote :

On 2008-11-18 11:52:08 -0500, Victor Bazarov <v.Abaza...@com Acast.netsaid:

Tonni Tielens wrote:

I'm trying to create a pure virtual class describing an
interface. Normally, when I do this I make the destructor
pure virtual so that, even if there are no members in the
class, it cannot be instantiated.

Why would you have an interface with no other members?
 Wouldn't it be pretty much useless as an interface?

It's a Java thing.

I don't think it's only Java. It's known as a tagging
interface, and it potentially has a role in any staticly typed
language which supports polymorphism. I think I've actually
used it once in C++; C++ usually has other ways of solving the
problem, however, which are generally preferred (because they
can be made to work with non class types as well). Even in the
standard, the iterator_tag hierarchy could be considered an
example of this.

--
James Kanze (GABI Software) email:james.kan ze@gmail.com
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