Add a method to an STL container?

[Joseph Turian]

Without modifying the STL code directly, what's the *simplest* way to
extend an STL container by adding a method to it?

e.g. I want to add method foo() to list<T>.

Can I do this without creating a list<T> subtype with a foo() method?
i.e. Can I add foo() directly to list<T>?

Joseph

[benben]

> Without modifying the STL code directly, what's the *simplest* way to

extend an STL container by adding a method to it?

e.g. I want to add method foo() to list<T>.

Can I do this without creating a list<T> subtype with a foo() method?
i.e. Can I add foo() directly to list<T>?

Joseph

namespace MyListOps
{
template <typename T>
void foo(std::list<T>& );

// ...
}

regards,
ben

[snnn]

Joseph Turian wrote:

Without modifying the STL code directly, what's the *simplest* way to
extend an STL container by adding a method to it?

e.g. I want to add method foo() to list<T>.

Can I do this without creating a list<T> subtype with a foo() method?
i.e. Can I add foo() directly to list<T>?

Joseph

define a new class is-a list<T> or has-a list<T>,no simpler way.

[Cy Edmunds]

"Joseph Turian" <tu****@gmail.com> wrote in message
news:11**********************@g14g2000cwa.googlegr oups.com...

Without modifying the STL code directly, what's the *simplest* way to
extend an STL container by adding a method to it?

e.g. I want to add method foo() to list<T>.

Can I do this without creating a list<T> subtype with a foo() method?
i.e. Can I add foo() directly to list<T>?

Joseph

It's almost certainly the wrong thing to do. Now you have a collection type
which is different from the standard type only because of some function you
wanted to throw in. If you decide later to add another function you will
have three "different" collection types! I strongly endorse Ben's solution
which is to make your convenience functions non-member functions. That
solution is extendable to as many convenience functions as you like without
ever making your own collection class.

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

[John Dibling]

benben wrote:

namespace MyListOps
{
template <typename T>
void foo(std::list<T>& );

// ...
}

regards,
ben

Another option:

template<class Iterator>
void foo(Iterator first, Iterator last);

This is independant of the container type, just like STL algorithms
are. In fact, it can be used on a C-style array, just like STL
algorithms can be.
Take care,

John Dibling

[Peter Julian]

"Joseph Turian" <tu****@gmail.com> wrote in message
news:11**********************@g14g2000cwa.googlegr oups.com...

Without modifying the STL code directly, what's the *simplest* way to
extend an STL container by adding a method to it?

e.g. I want to add method foo() to list<T>.

Can I do this without creating a list<T> subtype with a foo() method?
i.e. Can I add foo() directly to list<T>?

Joseph

Wrap the stl container in a class.

template< class T >
class SuperList
{
std::list<T> m_list;
public:
foo();
... // push, pop, etc
};

[msalters]

snnn wrote:

define a new class is-a list<T> or has-a list<T>,no simpler way.

You can't get 100% is-a:

std::auto_ptr<std::list<int> > will fail, as it tries
to call std::list::~list instead of ypur foolist::~foolist.

Regards,
Michiel Salters