Use static type from template pointer type

Max13

Hi,

Le title isn't really clear, but I didn't know how to write it, sorry if it's also a duplicate.

I'm subclassing QAbstractListModel as a template class named AbstractListModel which contains an internal QList (of pointer types). Obviously, my subclass is a template to define the QList content, but I also use it with QVariants.

I want the AbstractListModel subclasses to inherit from *AbstractListModel<MyClass > (which will set the QList content to <MyClass *> also) BUT I'm using the full type (not the pointer, the static type) to convert from QVariant, my problem is that I don't know how to write the full template type from the pointer one.

Example:
@template <typename T = int *>
void test()
{
T a = 0; // Here, a is a pointer to int.
??????? // And here I want to declare an int
}@

Is there a way to declare a static type if you know that T is a pointer type?

andreyc

Yes, there is a way in c++ 11, but I don't think it is very useful.
@
template <typename T = int *>
void test()
{
T a = 0; // Here, a is a pointer to int.
auto b = *a; // And here I want to declare an int
std::cout << b << std::endl; // core dump is here :-)
}
@

PS: Why not inverse it.
@
template <typename T = int>
void test()
{
T a = 0;
T* b = new T;
std::cout << b << std::endl; // core dump is here :-)
}
@

Max13

This is just an example. The type will never be an int but a class, and there will be no default type.

I wrote it that way only as a comprehensible example, so inversion is excluded (as the question is how to write the static type from the pointer type), as well as auto ;)

Thanks anyway

andreyc

Thank you for interesting question.
BTW why auto is excluded ?

I have another example. But it uses auto
@
/** http://www.cplusplus.com/reference/memory/pointer_traits/

g++ -std=c++11 -o test2 test2.cpp
*/
// pointer_traits example
#include <iostream>
#include <memory>

// using pointer_traits to determine return type:
template <class T>
typename std::pointer_traits<T>::element_type dereference_pointer (T pt) {
return *pt;
}

class A
{
public:
A() : m_number(0) {}
~A() {}

    void setNumber(int number) {
        m_number = number;
    }
    int number() const {
        return m_number;
    }

private:
    int m_number;

};

class B
{
public:
B() {}
~B() {}
};

int main(int, char**)
{
int* foo = new int(1);
std::shared_ptr<int> bar (new int(2));

std::cout << "foo: " << dereference_pointer (foo) << '\n';
std::cout << "bar: " << dereference_pointer (bar) << '\n';

delete foo;

A* pa = new A;
pa->setNumber(23);

auto a = dereference_pointer(pa);
a.setNumber(34);

std::cout << "pa = " << pa->number() << std::endl;
std::cout << "a = " << a.number() << std::endl;

B* pb = new B;
auto b = dereference_pointer(pb);
b.setNumber(34);

return 0;

}
@

Max13

[quote author="andreyc" date="1401232860"]BTW why auto is excluded ?
[/quote]
I wrongly thought auto was used to declare ints. So it's not excluded, my mistake.

But as I'm reading your post, I think it's still now what I'm asking as I don't think using the value is useful, maybe my question isn't clear.

Let's imagine I have this class:
@template <typename T>
class A
{
T m_dummy;
}@

Here, if I instantiate A with T = QWidget * (this is an example), m_dummy will be a pointer to QWidget.

If I instantiate A with T = QWidget, m_dummy will be of type QWidget (the plain type, not a pointer).

Now, my question is: How to write A to be able to instantiate it with T = QWidget * and to have m_dummy to be of type QWidget (the plain type, not the pointer, as is T) ?

Hope it's clearer.

andreyc

Are you sure that you need to declare a variable. Maybe you can use some getter function that will return type auto and compiler will deduce the type.
Could you post an example that reflects your design.

Max13

[quote author="andreyc" date="1401243568"]Could you post an example that reflects your design.[/quote]

What I previously wrote was part of what I need.

@template <typename T>
class A
{
private:
QList<T> m_list;
T m_tmp;
}@

Here, if I instantiate A with T = QWidget *, how to write m_tmp to be the plain type and not the pointer?

I wonder how to do it, today I have a solution: Instantiate A with plain type and QList<T>*, but I didn't know how to write it if T = pointer.

andreyc

[quote author="Max13" date="1401266922"]I wonder how to do it, today I have a solution: Instantiate A with plain type and QList<T>*, [/quote]
That I was trying to say when I talking about "inverse it" :-)
[quote author="Max13" date="1401266922"]but I didn't know how to write it if T = pointer.[/quote]I don't know it either. That is why I asked about getter function. With a function you can pass a pointer to a variable and a compiler can deduce the type using
@
template <class T>
typename std::pointer_traits<T>::element_type dereference_pointer (T pt) {
return *pt;
}
@

Anyway, I'm glad that you've resolved the issue.

Could you please prepend "[SOLVED]" to the title of your post. So the other good people will be able to find your solution.

Max13

[quote author="andreyc" date="1401268410"]Could you please prepend "[SOLVED]" to the title of your post.[/quote]

Actually it's not solved ^^'
My own problem is solved, but not "How to write it" :/

bipll

g++ 4.7.2 compiles it with -std=c++98, got no real idea whether it is in accordance with any standard:
@#include <iostream>

template<class T> struct Deref {
typedef T Refed;
};

template<typename T> struct Deref<T *> {
typedef T Refed;
};

template<typename T> void omg(T t) {
typename Deref<T>::Refed defer(0);
std::cout << defer << std::endl;
}

int main() {
omg((int *)NULL);
}@