using iterator in template function

[wuych]

I have a question about using iterator in template function

//*****code starts here*****************************
#include <vector>
using std::vector;

template<typename Tvoid foo( vector<T& a )
{
vector<T>::iterator i; // ERROR, can't use iterator
}

int main()
{
vector<inta;
foo(a);
}
//******code ends here ****************************
//****** compiler: gcc 4.1.3 ************************

Can anyone tell me why it's not allowed to use the iterator this way?
Maybe the best way is to use Iterator directly as STL does.
template<typename Itervoid foo(Iter b, Iter e);
Thank you.

[Pete Becker]

On 2007-11-08 11:10:26 -0500, wuych <ya********@gmail.comsaid:

I have a question about using iterator in template function

//*****code starts here*****************************
#include <vector>
using std::vector;

template<typename Tvoid foo( vector<T& a )
{
vector<T>::iterator i; // ERROR, can't use iterator
}

int main()
{
vector<inta;
foo(a);
}
//******code ends here ****************************
//****** compiler: gcc 4.1.3 ************************

Can anyone tell me why it's not allowed to use the iterator this way?

vector<T>::iterator is what's known as a "dependent name". Its meaning
might be different for different types T (although a program that
specialized vector<Whateverand made iterator the name of a data
object would be quite perverse). So the rule is that you have to tell
the compiler that vector<T>::iterator names a type:

typename vector<T>::iterator i; // okay

Maybe the best way is to use Iterator directly as STL does.
template<typename Itervoid foo(Iter b, Iter e);

Maybe. That way your algorithm can operate on any sequence, and not
just on a vector. That's what most of the standard library algorithms
do. But it depends on what that function is actually doing.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)