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Template functions and inheritance

 
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Gareth Stockwell
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PostPosted: Wed Dec 22, 2004 9:04 am    Post subject: Template functions and inheritance Reply with quote



Similar questions to this have been asked before, but so far I haven't
found a satisfactory answer, so...

Consider the following:


struct A { };
struct B : public A { };

template<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

template<> void foo<A>(const A&)
{ std::cout << "A" << std::endl; }

int main() {

foo(1); // prints generic
foo(A()); // prints A
foo(B()); // prints generic
}


The last call matches the unspecialised version of foo. Let us say
that this is not the behaviour I want - since B() is convertible to a
const A& reference, I want to make the last call resolve to the
specialised version.

With a bit of blood, sweat and template metaprogramming, this is
possible - just rewrite the foos as static member functions of a
template class, and then use a compile-time check on whether T
inherits from A, to select which version gets called:


template struct Foo
{ static void call(const T&) { std::cout << "generic" << std::endl; }
};

template struct Foo<T,true>
{ static void call(const T&) { std::cout << "A" << std::endl; } };

template void new_foo(const T& x)
{ Foo<T, boost::is_derived_and_base(x); }


The problem I have is that, if we now templatize the base class, I can
no longer see a way to force the correct resolution.


template<typename T> struct A { };
struct B : public A<int> { };

tempate<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

template { std::cout << "A" << std::endl; }


This problem reduces to determining whether a given type is derived
from *any instantiation* of a given template - which I don't think is
possible. But then maybe I'm thinking about it in the wrong way...

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Alberto Barbati
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PostPosted: Thu Dec 23, 2004 11:45 am    Post subject: Re: Template functions and inheritance Reply with quote



Gareth Stockwell wrote:
Quote:

This problem reduces to determining whether a given type is derived
from *any instantiation* of a given template - which I don't think is
possible. But then maybe I'm thinking about it in the wrong way...


See my reply to your other post and remove the enable_if I put there.

Alberto

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Graeme Prentice
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PostPosted: Thu Dec 23, 2004 3:14 pm    Post subject: Re: Template functions and inheritance Reply with quote



On 22 Dec 2004 04:04:21 -0500, Gareth Stockwell wrote:


[snip]

Quote:

template<typename T, bool IsDerivedFromA
struct Foo
{ static void call(const T&) { std::cout << "generic" << std::endl; }
};

template struct Foo { static void call(const T&) { std::cout << "A" << std::endl; } };

template void new_foo(const T& x)
{ Foo(x); }


I assume you meant to call the function called call here ...

Quote:


The problem I have is that, if we now templatize the base class, I can
no longer see a way to force the correct resolution.


template<typename T> struct A { };
struct B : public A<int> { };

tempate<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

template { std::cout << "A" << std::endl; }


This problem reduces to determining whether a given type is derived
from *any instantiation* of a given template - which I don't think is
possible. But then maybe I'm thinking about it in the wrong way...


Well if you don't need to consider private base classes; multiple
inheritance or single inheritance with more than one A hierarchy then you can do it as below - similar to the other code I
posted except now the first f1 function is preferred if a U* can
implicitly convert to a T<W>* because a standard conversion is better
than an elipsis conversion. The only way a U* can implicitly convert to
a T<W>* is if U is derived from T<W>. As I mentioned in my other post,
14.8.2.1/3 allows a pointer to derived to convert to pointer to base if
base is a template-id - however base has to be an unambiguous base
class and there has to be only one possibility for the deduction -
which means that not only can multiple inheritance cause the following
code to fail, it can also fail with single inheritance if U inherits an
A<W> which inherits an A<Z> because now the argument deduction doesn't
know whether to deduce to Z or W so will choose the elipsis conversion
and give the wrong answer. The only solution to this is to create
specializations that explicitly check for derivation from A<W> - the
known problem cases - I don't believe there's any general solution - and
no specific solution if the template argument (W) is a template
parameter with unlimited possibilities. I don't think you can even
produce a compile time error for these cases.

Graeme



#include <iostream>
#include <ostream>

template<typename X> struct ATemplateClass { };

typedef ATemplateClass<int> Is;
typedef int IsNot;

struct IsToo : public Is
{
};

template<typename U, template
struct IsDerivedFromAnInstantiationOf {

typedef char t1[1];
typedef char t2[2];

template <typename W>
static t1& f1( T<W> * );

static t2& f1(...);

static const bool result =
(sizeof(IsDerivedFromAnInstantiationOf::f1( (U*)0 )) == 1);
};

static const bool should_be_true =
IsDerivedFromAnInstantiationOf<Is, ATemplateClass>::result;

static const bool should_be_false =
IsDerivedFromAnInstantiationOf<IsNot, ATemplateClass>::result;

static const bool should_be_true_too =
IsDerivedFromAnInstantiationOf<IsToo, ATemplateClass>::result;

int main()
{
std::cout << should_be_true << ' '
<< should_be_false << ' '
<< should_be_true_too;
}




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Maxim Yegorushkin
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PostPosted: Thu Dec 23, 2004 5:01 pm    Post subject: Re: Template functions and inheritance Reply with quote

Gareth Stockwell wrote:

Quote:
Consider the following:
struct A { };
struct B : public A { };
template<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }
template<> void foo<A>(const A&)
{ std::cout << "A" << std::endl; }
int main() {
foo(1); // prints generic
foo(A()); // prints A
foo(B()); // prints generic
}

This problem can be solved using SFINAE and a little trick to make the
generic function be not always a best match. I think the following code is
easy to grasp, though it may be worth a couple comments:

#include
#include "boost/utility/enable_if.hpp"
#include "boost/type_traits/is_base_and_derived.hpp"

template<class T>
void foo(T const&, long) // (1)
{
std::cout << "genericn";
}

struct A {};
struct B : A {};
struct C {};

void foo(A const&, int) // (2)
{
std::cout << "A overloadn";
}

// A's successor overload
template typename boost::enable_if<boost::is_base_and_derived::type
foo(T const& t, int) // (2)
{
return foo(static_cast<A const&>(t), 0);
}

int main(int ac, char** av)
{
A a;
B b;
C c;

foo(a, 0);
foo(b, 0);
foo(c, 0);
}

Outputs:

A overload
A overload
generic

(1) We always call foo with a dummy zero parameter, whose best match is
int. Our generic function takes it as long, thus requiring an integral
conversion.
(2) We make our specializations take the dummy integer as int. This allows
to override the generic template with another template. The fact that
another template accepts an int, rather then long, makes it a better match
then the generic template, because it does not require an integral
conversion, resulting in no ambiguity between the templates that would
arise if there was not the dummy parameter, or both templates accepted it
as the same integer type.

I like this technique because it allows one to keep specializing a
template by other templates without having to rewrite the first template.

--
Maxim Yegorushkin

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dieterbeaven@yahoo.com.au
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PostPosted: Thu Dec 23, 2004 9:14 pm    Post subject: Re: Template functions and inheritance Reply with quote

Hi Gareth,

One way to solve your issue whilst still using templates is to force
the instanciation of foo<A> for B():

int main()
{ foo(1); // prints generic
foo(A()); // prints A
foo<A>(B()); // prints A as required
}

However, if the design you have suggests B 'is-a' A, then an approach
that uses the principle of 'programming to interface' ( i.e. Design
Patterns - GOF ) might be suggested. In that case the client code
should only ever use pointers or references to A, then foo<A> is the
only instanciation that would ever naturally occur in client code.
Regards,

Dieter


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Allen Yao
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PostPosted: Thu Dec 23, 2004 9:40 pm    Post subject: Re: Template functions and inheritance Reply with quote

I thought out a solution. The main idea is :
(1) derive class template A from a unique tag class named ATag;
(2) define an type named ATemplateParameter to record A's template
parameter.

The solution also based on the SFINAE mechanism (boost::enable_if).

#include <iostream>
#include <boost/utility/enable_if.hpp>
#include <boost/type_traits.hpp>

struct ATag {};

template <typename T>
struct A : public ATag
{
typedef T ATemplateParameter;
};

struct B : public A<int>
{};

template <typename T>
typename boost::disable_if<boost::is_base_and_derived::type
foo(const T& t)
{ std::cout << "generic" << std::endl; }

template void foo_a(const A<typename T::ATemplateParameter>&)
{ std::cout << "A
template typename boost::enable_if<boost::is_base_and_derived::type
foo(const T& t)
{ foo_a<T>(t); }

int main()
{
int i;
A<int> a;
B b;

foo(i);
foo(a);
foo(b);
}

The output is:

generic
A<T>
A<T>

Note that we didn't consider the rare situation where
B derives from two different instatiations of template A, for example

struct B : public A<int>, public A<double> { ... };

which forms the diamond inheritance structure.

"Gareth Stockwell" <gareth_stockwell (AT) hotmail (DOT) com>
news:172ce8bd.0412210755.7469e974 (AT) posting (DOT) google.com...
Quote:
Similar questions to this have been asked before, but so far I haven't
found a satisfactory answer, so...

Consider the following:


struct A { };
struct B : public A { };

template<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

template<> void foo<A>(const A&)
{ std::cout << "A" << std::endl; }

int main() {

foo(1); // prints generic
foo(A()); // prints A
foo(B()); // prints generic
}


The last call matches the unspecialised version of foo. Let us say
that this is not the behaviour I want - since B() is convertible to a
const A& reference, I want to make the last call resolve to the
specialised version.

With a bit of blood, sweat and template metaprogramming, this is
possible - just rewrite the foos as static member functions of a
template class, and then use a compile-time check on whether T
inherits from A, to select which version gets called:


template struct Foo
{ static void call(const T&) { std::cout << "generic" << std::endl; }
};

template struct Foo { static void call(const T&) { std::cout << "A" << std::endl; } };

template void new_foo(const T& x)
{ Foo(x); }


The problem I have is that, if we now templatize the base class, I can
no longer see a way to force the correct resolution.

template<typename T> struct A { };
struct B : public A<int> { };

tempate<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

template { std::cout << "A" << std::endl; }

This problem reduces to determining whether a given type is derived
from *any instantiation* of a given template - which I don't think is
possible. But then maybe I'm thinking about it in the wrong way...



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dave
Guest





PostPosted: Thu Dec 23, 2004 9:42 pm    Post subject: Re: Template functions and inheritance Reply with quote

Your first problem is easy to solve if you don't insist on using
specialization:

------ schnipp ------
#include <boost/utility/enable_if.hpp>
#include <boost/type_traits/is_convertible.hpp>

struct A { };
struct B : public A { };

template<typename T>
typename boost::disable_if<boost::is_convertible::type
foo(T const&);

char* foo(A const&);

int main() {

int x = foo(1); // first overload
char* y = foo(A()); // second overload
char* z = foo(B()); // second overload
}
------ schnapp ------

Secondly, determining that a type is derived from any specialization of
a given class template is easy:

----- schnippy -----
#include <boost/mpl/bool.hpp>

template <class T, template
struct is_derived_from_or_same_as_specialization_of
{
typedef char yes;
typedef char(&no)[2];

template <class U>
static yes test(X<U>*);
static no test(...);

static bool const value = sizeof(test((T*)0)) == sizeof(yes);
typedef boost::mpl::bool_<value> type;
};

template <class T, template
struct is_derived_from_or_same_as_specialization_of<T&,X>
: boost::mpl::false_ {};

template <class> struct A {};
template <class T> struct B : A<T> {};

int a1[!is_derived_from_or_same_as_specialization_of<int,A>::value];
int a2[is_derived_from_or_same_as_specialization_of<A::value];
int a3[is_derived_from_or_same_as_specialization_of<B::value];
#include <boost/mpl/bool.hpp>

template <class T, template
struct is_derived_from_or_same_as_specialization_of
{
typedef char yes;
typedef char(&no)[2];

template <class U>
static yes test(X<U>*);
static no test(...);

static bool const value = sizeof(test((T*)0)) == sizeof(yes);
typedef boost::mpl::bool_<value> type;
};

template <class T, template
struct is_derived_from_or_same_as_specialization_of<T&,X>
: boost::mpl::false_ {};

template <class> struct A {};
template <class T> struct B : A<T> {};

int a1[!is_derived_from_or_same_as_specialization_of<int,A>::value];
int a2[is_derived_from_or_same_as_specialization_of<A::value];
int a3[is_derived_from_or_same_as_specialization_of<B::value];
------ schnappy ------

Caveat: the implementation above will barf if A<T> is a private base
class. See Rani Sharoni's is_base_and_derived trick in the boost
implementation of that trait for a way to get around that problem.

Cheers,
--
Dave Abrahams
Boost Consulting
http://www.boost-consulting.com


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dave_abrahams
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PostPosted: Sat Dec 25, 2004 10:07 am    Post subject: Re: Template functions and inheritance Reply with quote

My second example looks more complicated than it really is. Somehow I
pasted two copies of the same code there ;-/. Sorry.
--
Dave Abrahams
Boost Consulting
http://www.boost-consulting.com


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dave_abrahams
Guest





PostPosted: Sat Dec 25, 2004 8:56 pm    Post subject: Re: Template functions and inheritance Reply with quote

and season well.
Place in a large bowl and cool.
Add seasoned breadcrumbs and a little of the tomato gravy,
enough to make the mixture pliable.
Divide the stuffing among the cabbage leaves then roll.
Place seam down in a baking pan.
Ladle tomato gravy on top,
and bake at 325° for 30 - 45 minutes.



Umbilical Cordon Bleu

Nothing is so beautiful as the bond between mother and child,
so why not consume it?
Children or chicken breasts will work wonderfully also.

4 whole umbilical chords (or baby breasts, or chicken breasts)
4 thin slices of smoked ham, and Gruyere cheese
Flour
eggwash (milk and eggs)
seasoned bread crumbs
1 onion
minced
salt
pepper
butter
olive oil

Pound the breasts flat (parboil first if using umbilical
cords so they won?t be tough).
Place a slice of ham and cheese on each, along with some minced onion
then fold in half, trimming neatly.
Dredge in flour, eggwash, then seasoned breadcrumbs;
allow to sit for a few minutes.
Sauté in butter and olive oil until golden brown,
about 6 minutes on each side.



Shish Kababes

As old as the hills, this technique has employed seafood, beef, pork, lamb,
poultry, and vegetables; just about anything can be grilled, and young humans
are no exception!

High quality marinade (Teriyaki and garlic perhaps)
1 inch cubes of tender meat, preferably from the nursery
Onions
bell peppers
Wooden or metal skewers

Marinate the meat overnight.
Get the grill good and hot while placing meat, vegetables, and
fruit such as pineapples or cherries on the skewers.
Don?t be afraid to u


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dave_abrahams
Guest





PostPosted: Sat Dec 25, 2004 10:29 pm    Post subject: Re: Template functions and inheritance Reply with quote

thoroughly cleaned and washed).
Smother slowly for at least 2 hours, adding small amounts of water
when it starts to stick.
Stir frequently.
When ready - serve with rice, grilled smoked sausage, green salad, and iced tea.
Coffee and apple pie then brandy.



Maternity Ward Pot Luck Dinner

If you can?t get anything fresh from the hospital, nursery, or morgue;
you can at least get rid of all the leftovers in your refrigerator.

1 - 2 lbs. cubed meat (human flesh, chicken, turkey, beef...)
1 -2 lbs. coarsely chopped vegetables
(carrots, potatoes, turnips, cauliflower, cabbage...)
Bell pepper
onions
garlic
ginger
salt pepper, etc.
Olive oil
butter

Brown the meat and some chopped onions, peppers, and garilic in olive oil,
place in baking dish, layer with vegetables seasoning and butter.
Bake at 325° for 30 - 45 minutes.
Serve with hot dinner rolls, fruit salad and sparkling water.



Bébé Buffet 1

Show off with whole roasted children replete with apples in mouths -
and babies? heads stuffed with wild rice. Or keep it simple with a
hearty main course such as stew, lasagna, or meat loaf.

Some suggestions

Pre-mie pot pies, beef stew, leg of lamb, stuffed chicken, roast pork spiral ham,
Cranberry pineapple salad, sweet potatoes in butter, vegetable platter, tossed salad with tomato and avocado, parsley new potatoes, spinich cucumber salad, fruit salad
Bran muffins, dinner rolls, soft breadsticks, rice pilaf, croissants
Apple cake with rum sauce, frosted banana nut bread sherbet, home made brownies
Iced tea, water, beer, bloody marys, lemonade, coffee

The guests select food, beverages, silverware... everything from the buffet table.
They move to wherever they are


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Allan W
Guest





PostPosted: Tue Dec 28, 2004 10:39 am    Post subject: Re: Template functions and inheritance Reply with quote

I think that templates are the wrong solution to this problem.


struct A { };
struct B : public A { };

template<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

// Not a template
void foo(const A&)
{ std::cout << "A" << std::endl; }

int main() {
foo(1); // prints generic
foo(A()); // prints A
foo(B()); // prints generic!
}


I think that ought to work... but it does not. Am I reading 13.3.3/1
wrong?
....a viable function F1 is defined to be a better function than another
viable function F2 if...
....F1 is a non-template function and F2 is a template function
specialization...


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Thomas Mang
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PostPosted: Tue Dec 28, 2004 4:07 pm    Post subject: Re: Template functions and inheritance Reply with quote


"Allan W" <allan_w (AT) my-dejanews (DOT) com> schrieb im Newsbeitrag
news:1104182618.066391.88350 (AT) f14g2000cwb (DOT) googlegroups.com...
Quote:
I think that templates are the wrong solution to this problem.


struct A { };
struct B : public A { };

template<typename T> void foo(const T&)
{ std::cout << "generic" << std::endl; }

// Not a template
void foo(const A&)
{ std::cout << "A" << std::endl; }

int main() {
foo(1); // prints generic
foo(A()); // prints A
foo(B()); // prints generic!
}


I think that ought to work... but it does not. Am I reading 13.3.3/1
wrong?
...a viable function F1 is defined to be a better function than another
viable function F2 if...
...F1 is a non-template function and F2 is a template function
specialization...


See 13.3.3.1.4/1

The template function is called because it's an identity conversion (exact
match). The overloaded function requires a derived -> base conversion
(Conversion). And exact match is better than Conversion.


Thomas



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Thomas Mang
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PostPosted: Tue Dec 28, 2004 10:19 pm    Post subject: Re: Template functions and inheritance Reply with quote

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Allan W
Guest





PostPosted: Wed Dec 29, 2004 1:53 am    Post subject: Re: Template functions and inheritance Reply with quote

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