Why typename can not be inherented from base class?

[PengYu.UT]

Hi,

I'm wondering why typename can not be inherented from base class. An
example is shown below. If I have to typedef value_type anyway, I think
it would be easier to define the 5 types in iterator like "typedef T
value_type" without inherent from std::iterotor.

Thanks,
Peng

#include <iterator>

template <typename T>
class iterator : public std::iterator<std::random_access_iterator_tag,
T> {
public:
typedef typename std::iterator<std::random_access_iterator_tag,
T>::value_type value_type;//this line can not be deleted
private:
value_type v;
};

int main(){
iterator<int> it;
}

 

[Victor Bazarov]

I'm wondering why typename can not be inherented from base class. An

I am not sure what your question is. What can't be inherited?

template<class T> class Foo {
public:
typedef T* pointer;
};

template<class T> class Bar : public Foo<T> {
public:
typedef typename Foo<T>:ointer pointer; // typedef inherited
private:
pointer ptr;
};

example is shown below. If I have to typedef value_type anyway, I
think it would be easier to define the 5 types in iterator like
"typedef T value_type" without inherent from std::iterotor.

Easier? In what way?

Thanks,
Peng

#include <iterator>

template <typename T>
class iterator : public std::iterator<std::random_access_iterator_tag,
T> {
public:
typedef typename std::iterator<std::random_access_iterator_tag,
T> value_type value_type;//this line can not be deleted
private:
value_type v;
};

int main(){
iterator<int> it;
}

V

 

[Greg Comeau]

I'm wondering why typename can not be inherented from base class. An
example is shown below. If I have to typedef value_type anyway, I think
it would be easier to define the 5 types in iterator like "typedef T
value_type" without inherent from std::iterotor.

#include <iterator>

template <typename T>
class iterator : public std::iterator<std::random_access_iterator_tag,
T> {
public:
typedef typename std::iterator<std::random_access_iterator_tag,
T>::value_type value_type;//this line can not be deleted
private:
value_type v;
};

int main(){
iterator<int> it;
}

The typename is not being singled out, as there is a more general issue:
it has to do with names dependent upon the template parameter. See
http://www.comeaucomputing.com/techtalk/templates/whymembernotfound

 

[PengYu.UT]

The typename is not being singled out, as there is a more general issue:
it has to do with names dependent upon the template parameter. See
http://www.comeaucomputing.com/techtalk/templates/whymembernotfound

Sorry, I can't reach this link. Would you please check what is wrong?

Thanks,
Peng

 

[Greg Comeau]

Sorry, I can't reach this link. Would you please check what is wrong?

http://www.comeaucomputing.com/techtalk/templates/#whymembernotfound

 

[PengYu.UT]

I am not sure what your question is. What can't be inherited?

template<class T> class Foo {
public:
typedef T* pointer;
};

template<class T> class Bar : public Foo<T> {
public:
typedef typename Foo<T>:ointer pointer; // typedef inherited

Why can't I simple delete the above line?

private:
pointer ptr;
};


Easier? In what way?

Since "pointer" in the base class is defined from T anyway, why can't I
define "pointer" in iterator(my example) without inherent the base
class std::iterator.

 

[Victor Bazarov]

Why can't I simple delete the above line?

You can't? What gives? Have you tried just deleting it?
^^^^^^^^^^^^
And if you delete it, what happens to this line? You may need to change
it, don't you think?

Since "pointer" in the base class is defined from T anyway, why can't I
define "pointer" in iterator(my example) without inherent the base
class std::iterator.

Can't you?

V