Returning iterator from template class

[jois.de.vivre]

Hi All,

I'm trying to write a wrapper class for std::vector to extend some of
its functionality. The problem I'm getting into is returning an
iterator type from a member function. Here is the essense of what I'm
trying to do:

//-------code-------

#include <vector>
using namespace std;

template <class T>
class ExtendedVector
{
private:
vector<T> vals;
public:
vector<T>::iter ator begin()
{
return vals.begin();
}
};

//-------end code-------

However, this gives me the error

error: expected ';' before 'begin'

What's the proper way to achieve what I'm trying to do?

[Noah Roberts]

jois.de.vi...@g mail.com wrote:

Implement thus:

template <class T>
class ExtendedVector
{
private:
vector<T> vals;
public: // the following not only saves typing but allows you to
implement
// with a different object type other than vector...just
change the defs here
// and clients will fall in line so long as they don't try
to bypass your interface.

typedef typename vector<T>::iter ator iterator_t;
typedef typename vector<T>::cons t_iterator const_iterator_ t;
iterator_t begin() // changed return type to above typedefs
{
return vals.begin();
} // implement for const objects...
const_iterator_ t begin() const { return vals.begin(); } };

//-------end code-------

ExtendedVector< T> t;
ExtendedVector< T>::iterator_t it;

[Sumit Rajan]

<jo***********@ gmail.com> wrote in message
news:11******** *************@g 10g2000cwb.goog legroups.com...

Hi All,

I'm trying to write a wrapper class for std::vector to extend some of
its functionality. The problem I'm getting into is returning an
iterator type from a member function. Here is the essense of what I'm
trying to do:

//-------code-------

#include <vector>
using namespace std;

template <class T>
class ExtendedVector
{
private:
vector<T> vals;
public:
vector<T>::iter ator begin()

typename vector<T>::iter ator begin()

Regards,

Sumit.
--
Sumit Rajan <su*********@gm ail.com>

[jois.de.vivre]

>

typedef typename vector<T>::iter ator iterator_t;
typedef typename vector<T>::cons t_iterator const_iterator_ t;

iterator_t begin() // changed return type to above typedefs
{
return vals.begin();
}

// implement for const objects...
const_iterator_ t begin() const { return vals.begin(); }

Wonderful! I guess my next question is, how did you know to do that? At
this point I don't even know what the 'typedef typename' part of the
above code snippet means. More specifically I suppose is, where can I
get a detailed discussion about templates. Any recommendations on
books? web sites?

[dan2online]

jo***********@g mail.com wrote:

Hi All,

I'm trying to write a wrapper class for std::vector to extend some of
its functionality. The problem I'm getting into is returning an
iterator type from a member function. Here is the essense of what I'm
trying to do:

//-------code-------

#include <vector>
using namespace std;

template <class T>
class ExtendedVector
{
private:
vector<T> vals;
public:
vector<T>::iter ator begin()
{
return vals.begin();
}
};

//-------end code-------

However, this gives me the error

error: expected ';' before 'begin'

What's the proper way to achieve what I'm trying to do?

You need to point out that the vector<T>::iter ator is typename.
If you want to extend the vector, inheritance is better than
composition in your case.

[Sumit Rajan]

<jo***********@ gmail.com> wrote in message
news:11******** *************@i 39g2000cwa.goog legroups.com...

get a detailed discussion about templates. Any recommendations on
books? web sites?

Book Recommendation:
"C++ Templates" by David Vandevoorde, Nicolai M. Josuttis.

Regards,
Sumit.
--
Sumit Rajan <su*********@gm ail.com>

[Marcus Kwok]

jo***********@g mail.com wrote:

Wonderful! I guess my next question is, how did you know to do that? At
this point I don't even know what the 'typedef typename' part of the
above code snippet means. More specifically I suppose is, where can I
get a detailed discussion about templates. Any recommendations on
books? web sites?

Basically, the "typename" keyword is used for what are called "dependant
names" in a template, which tells the compiler that what follows it is
the name of a type and not e.g. a static object of a class.

As always, the FAQ is a good starting point:
http://www.parashift.com/c++-faq-lite/

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

[Noah Roberts]

jo***********@g mail.com wrote:

typedef typename vector<T>::iter ator iterator_t;
typedef typename vector<T>::cons t_iterator const_iterator_ t;

iterator_t begin() // changed return type to above typedefs
{
return vals.begin();
}

// implement for const objects...
const_iterator_ t begin() const { return vals.begin(); }

Wonderful! I guess my next question is, how did you know to do that? At
this point I don't even know what the 'typedef typename' part of the
above code snippet means. More specifically I suppose is, where can I
get a detailed discussion about templates. Any recommendations on
books? web sites?

Markus answered the question but it may have been a bit confusing.
Without the 'typename' keyword the compiler will think you are trying
to reference a value within the class:

class X
{
public: enum { v };
};

int val = X::v;

is more along the lines of what it thinks you mean because it can't
actually get complete information about the template at this point and
doesn't know what you are talking about. If it looks like a variable
it is unless it isn't....and right now it doesn't know that it isn't.

So you explicitly say, "The next name I use is a type name, not a
variable or value."

Really you are seing this:

typedef type new_name;

where type is "typename temp<type>::int ernal_type".

I did the typedef to abstract the internal representation of the class
so that if vector becomes the wrong container it can be changed without
affecting clients. They will have to be recompiled of course, but you
won't have to do a bunch of editing outside the class to reflect
changes done to its insides. You could even return your own iterator
that might be a simple pointer at some point. It isn't necissary to do
the typedefs, but it is good practice. All clients need to know is
that something that has the interface of an iterator will be returned
from those functions...the y don't need to know exactly what
implementation of an iterator they are getting (although it would be
important to know forward/reverse/random/insert...etc... more may be
necissary to truely encapsulate your type).

You can also read Bruce Eckel's Thinking in C++ V2 online and this
explains what typename is and where it needs to be used.

[Noah Roberts]

dan2online wrote:

You need to point out that the vector<T>::iter ator is typename.
If you want to extend the vector, inheritance is better than
composition in your case.

As a rule inheritance should be used only when necissary. It may well
be the case here but nothing in the OP indicates that.

[Jack Saalweachter]

dan2online wrote:

If you want to extend the vector, inheritance is better than
composition in your case.

Eehhhhh.

It's 'riskily dangerous' to inherit from vector<T> (or any other
standard containers) because they don't have virtual destructors. No
virtual destructors mean that if you ever end up storing your
super_vector<T> as a vector<T>*, and then delete that pointer, /the
universe explodes/. Or near enough.

There are few 'never's in programming, but as a rule of thumb, I have to
stop and think real hard about extending a class with no virtual destructor.
Jack Saalweachter