I somehow understand why member function templates cannot be virtual,
but my question is how I could achieve something similar to this:
class A {
public:
template<typena me Tvirtual f() = 0;
}
class B : public A {
public:
template<typena me Tf();
}
class C : public A {
public:
template<typena me Tf();
}
That is, I would like to force subclasses to implement a templated
member function f. 5 2963
dj wrote:
I somehow understand why member function templates cannot be virtual,
but my question is how I could achieve something similar to this:
class A {
public:
template<typena me Tvirtual f() = 0;
}
class B : public A {
public:
template<typena me Tf();
}
That is, I would like to force subclasses to implement a templated
member function f.
Forcing subclasses isn't the problem. How are you going to convince
the compiler to instantiate B::f<int>? The compiler only sees a call to
A::f<int>.
HTH,
Michiel Salters Mi************* @tomtom.com wrote:
dj wrote:
>I somehow understand why member function templates cannot be virtual, but my question is how I could achieve something similar to this:
class A { public: template<typena me Tvirtual f() = 0; }
class B : public A { public: template<typena me Tf(); }
That is, I would like to force subclasses to implement a templated member function f.
Forcing subclasses isn't the problem. How are you going to convince
the compiler to instantiate B::f<int>? The compiler only sees a call to
A::f<int>.
HTH,
Michiel Salters
I think I understand what you are saying. If I use:
B b;
A* pb = &b;
pb->f<int>();
then the compiler would instantiate A::f<int>, but not B::f<int>. If f
is pure this would be illegal anyway, so that is probably why templated
virtuals are not allowed in the first place.
However, let me repeat my question - how could I force the subclasses to
implement some function f (which is what i use the pure virtual for),
which itself should be templated? I could declare all possible function
prototypes instead of using a template, but is there a more elegant
solution?
dj wrote:
However, let me repeat my question - how could I force the subclasses to
implement some function f (which is what i use the pure virtual for),
which itself should be templated? I could declare all possible function
prototypes instead of using a template, but is there a more elegant
solution?
struct Base
{
template <class DerivedT, class T>
void foo()
{
DerivedT* derived( dynamic_cast<De rivedT*>(this) );
derived->template f<T>();
}
virtual ~Base(){}
};
struct Derived : Base
{
template <class T>
void f(){ ; }
};
void test_a()
{
Derived d;
Base& b( d );
b.foo<Derived, int>();
}
Derived is forced to implement f which is itself templated.... I don't
know whether this is usefull, but that seems to do what you ask.
Interesting syntax... Compiles on Comeau.
Regards,
W
werasm wrote:
dj wrote:
>However, let me repeat my question - how could I force the subclasses to implement some function f (which is what i use the pure virtual for), which itself should be templated? I could declare all possible function prototypes instead of using a template, but is there a more elegant solution?
struct Base
{
template <class DerivedT, class T>
void foo()
{
DerivedT* derived( dynamic_cast<De rivedT*>(this) );
derived->template f<T>();
}
virtual ~Base(){}
};
struct Derived : Base
{
template <class T>
void f(){ ; }
};
void test_a()
{
Derived d;
Base& b( d );
b.foo<Derived, int>();
}
Derived is forced to implement f which is itself templated.... I don't
know whether this is usefull, but that seems to do what you ask.
Interesting syntax... Compiles on Comeau.
Regards,
W
This code really does the thing, thank you. I agree it is very weird,
though. Especially the call "derived->template f<T>();" is something new
to me. I tried and it works as "derived->f<T>();", too.
What I need such uncommon construction for is to have many different
algorithms implemented in different subclasses of some base class. The
templated function f would return the formatted result of any algorithm
in different possible data types. I am sure there are many other ways to
achieve that but this one seemed straightforward to me. Obviously I was
mistaken.
dj wrote:
This code really does the thing, thank you. I agree it is very weird,
though. Especially the call "derived->template f<T>();" is something new
to me. I tried and it works as "derived->f<T>();", too.
The other way is technically correct. If the compiler compiles
derived->f<T>(), which VC++7.1 does do, its not iaw. standard. The
template syntax is for the compiler to discern that < is not a
relational operator, I think. Someone else can perhaps comment (my time
is limited right now).
Regards,
Werner This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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