Hi there,
I am learning template programming and there is a problem about
traits. Now consider a container and an iterator. Here is the code
// tag for const iterator and non-const iterator
struct non_const_itera tor_tag {};
struct const_iterator_ tag {};
// traits
template <typename T,
typename tag=non_const_i terator_tag>
struct Iter_traits
{
typedef random_access_i terator_tag iterator_catego ry;
typedef T value_type;
typedef ptrdiff_t difference_type ;
typedef T* pointer;
typedef T& reference;
};
// For const iterator ...
template <typename T>
struct Iter_traits<T, const_iterator_ tag>
{
typedef random_access_i terator_tag iterator_catego ry;
typedef T value_type;
typedef ptrdiff_t difference_type ;
typedef const T* pointer;
typedef const T& reference;
};
//Iter is an iterator
template <typename T, typename tag=non_const_i terator_tag>
class Iter
{
typedef typename Iter_traits<T, tag>::value_typ e value_type;
typedef typename Iter_traits<T, tag>::reference reference;
public:
Iter(T *c) :cont(c) {...}
// copy constructor to shift the non-const iterator to const
iterator
Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c)
{...}
private:
T *cont;
};
// Container is a container.
template <typename T>
class Container
{
...
public:
typedef Iter<T> iterator;
typedef Iter<T, const_iterator_ tag> const_iterator;
iterator begin() {...};
const_iterator begin() const {...};
iterator end() {...};
const_iterator end() const {...};
private:
T *data;
...
};
int main(void)
{
Container<int> c;
// copy constructor should be actived here
Container<int>: :const_iterator cit=c.begin();
return 0;
}
An error occurs while compliation.
`int* Iter<int, non_const_itera tor_tag>::cont' is private
I don't know why. It seems that it's due to the copy constructor.
Please help!
BTW, if I define begin() outside the class. e.g.
template <typename T>
Container<T>::i terator Container<T>::b egin(void)
{
...
}
The compiler will give a warning:
`typename Container<T>::i terator' is implicitly a typename
mplicit typename is deprecated, please see the documentation for
details
Is that anything wrong? Should I do something to prevent that warning?
Thanks in advance. 5 2394
"Rex_chaos" <re*******@21cn .com> wrote... I am learning template programming and there is a problem about traits. Now consider a container and an iterator. Here is the code
// tag for const iterator and non-const iterator struct non_const_itera tor_tag {}; struct const_iterator_ tag {};
// traits template <typename T, typename tag=non_const_i terator_tag> struct Iter_traits { typedef random_access_i terator_tag iterator_catego ry; typedef T value_type; typedef ptrdiff_t difference_type ; typedef T* pointer; typedef T& reference; };
// For const iterator ... template <typename T> struct Iter_traits<T, const_iterator_ tag> { typedef random_access_i terator_tag iterator_catego ry; typedef T value_type; typedef ptrdiff_t difference_type ; typedef const T* pointer; typedef const T& reference; };
//Iter is an iterator
template <typename T, typename tag=non_const_i terator_tag> class Iter { typedef typename Iter_traits<T, tag>::value_typ e value_type; typedef typename Iter_traits<T, tag>::reference reference; public: Iter(T *c) :cont(c) {...}
// copy constructor to shift the non-const iterator to const iterator
Why do you say "to shift the ..."? It's just a copy constructor.
Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c) {...}
private: T *cont; };
// Container is a container.
The comment above seems unnecessary. The class name is self-
documenting. template <typename T> class Container { ... public: typedef Iter<T> iterator; typedef Iter<T, const_iterator_ tag> const_iterator;
iterator begin() {...}; const_iterator begin() const {...}; iterator end() {...}; const_iterator end() const {...};
Drop the semicolons after the closing curly braces. The code
looks dirty due to them.
private: T *data; ... };
int main(void) { Container<int> c;
// copy constructor should be actived here Container<int>: :const_iterator cit=c.begin();
Why should it? c.begin() returns 'Container<int> ::iterator'.
You want to assign it to 'Container<int> ::const_iterato r'.
Those two are different types.
return 0; }
An error occurs while compliation.
`int* Iter<int, non_const_itera tor_tag>::cont' is private
I don't know why. It seems that it's due to the copy constructor.
No. It's due to the fact that you don't have a constructor that
constructs a const_iterator from a "regular" iterator.
Please help!
Define a parameterised constructor that would convert 'iterator'
into 'const_iterator '.
BTW, if I define begin() outside the class. e.g.
template <typename T> Container<T>::i terator Container<T>::b egin(void) { ... }
The compiler will give a warning: `typename Container<T>::i terator' is implicitly a typename mplicit typename is deprecated, please see the documentation for details
Is that anything wrong? Should I do something to prevent that warning?
You should declare such function this way:
template<typena me T>
typename
Container<T>::i terator Container<T>::b egin()
{
...
Victor
On 23 Oct 2003 13:22:26 -0700, Rex_chaos <re*******@21cn .com> wrote: Hi there,
//Iter is an iterator
template <typename T, typename tag=non_const_i terator_tag> class Iter { typedef typename Iter_traits<T, tag>::value_typ e value_type; typedef typename Iter_traits<T, tag>::reference reference; public: Iter(T *c) :cont(c) {...}
// copy constructor to shift the non-const iterator to const iterator Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c) {...}
this part looks suspicious. I don't know what it.c is but it is probably
private and
const_iterator do not have access to it . private: T *cont; };
// Container is a container.
I don't know why. It seems that it's due to the copy constructor. Please help!
BTW, if I define begin() outside the class. e.g.
template <typename T> Container<T>::i terator Container<T>::b egin(void) { ... }
The compiler will give a warning: `typename Container<T>::i terator' is implicitly a typename mplicit typename is deprecated, please see the documentation for details
Is that anything wrong? Should I do something to prevent that warning?
Thanks in advance.
--
grzegorz
> Why do you say "to shift the ..."? It's just a copy constructor.
Yes. But the copy constructor is not a common one, because the
incomming class is a non_const_itera tor wihile the destination class
is a const_iterator Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c) {...}
private: T *cont; }; template <typename T> class Container { ... public: typedef Iter<T> iterator; typedef Iter<T, const_iterator_ tag> const_iterator;
iterator begin() {...}; const_iterator begin() const {...}; iterator end() {...}; const_iterator end() const {...}; Drop the semicolons after the closing curly braces. The code looks dirty due to them.
Sorry, I forgot to erase the semicolons.
private: T *data; ... };
int main(void) { Container<int> c;
// copy constructor should be actived here Container<int>: :const_iterator cit=c.begin();
Why should it? c.begin() returns 'Container<int> ::iterator'. You want to assign it to 'Container<int> ::const_iterato r'. Those two are different types.
Take a look at the Container<T> class, I have two functions for
begin(), one returns the iterator and other returns the
const_iterator. When I call
Container<int>: :const_iterator cit=c.begin();
I hope the later begin() will be called. However, it won't. So I have
a copy constructor in my code.
No. It's due to the fact that you don't have a constructor that constructs a const_iterator from a "regular" iterator.
But I have a copy constructor for this purpose.
> > template <typename T, typename tag=non_const_i terator_tag> class Iter { typedef typename Iter_traits<T, tag>::value_typ e value_type; typedef typename Iter_traits<T, tag>::reference reference; public: Iter(T *c) :cont(c) {...}
// copy constructor to shift the non-const iterator to const iterator Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c) {...}
this part looks suspicious. I don't know what it.c is but it is probably private and const_iterator do not have access to it .
Try this code, and you see that in (copy) constructor, the class
itself can also access the private data.
class Test
{
public:
Test() {...}
Test(const Test& t) :data(t.data) {}
private:
int data;
};
"Rex_chaos" <re*******@21cn .com> wrote... template <typename T, typename tag=non_const_i terator_tag> class Iter { typedef typename Iter_traits<T, tag>::value_typ e value_type; typedef typename Iter_traits<T, tag>::reference reference; public: Iter(T *c) :cont(c) {...}
// copy constructor to shift the non-const iterator to const iterator Iter(const Iter<T, non_const_itera tor_tag>& it) :cont(it.c) {...}
this part looks suspicious. I don't know what it.c is but it is probably private and const_iterator do not have access to it .
Try this code, and you see that in (copy) constructor, the class itself can also access the private data.
class Test { public: Test() {...} Test(const Test& t) :data(t.data) {} private: int data; };
In this code, in the copy c-tor, 't' and '*this' are of the
_same_ type. In your attempt, iterator and const_iterator
are not even related.
Victor This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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