STL::Accessing the iterator for a "passed" container type

Hi,

I am passing a container type (e.g. list<int>) to a function template
but I am getting a compiler error when I try to declare an iterator
for that type in the function. However, I am able to invoke begin()
and end() without any problem. Is there a way to declare an iterator
for the container type? Sample code follows.

Thanks,
Ramesh

//////////////////////////////////////////////////////
//
///////////////////////////////////////////////////////
#include <iostream>
#include <list>

using namespace std;

//////////////////////////////////////////////////////
//
///////////////////////////////////////////////////////
template<typename T>
void
access_the_iterator(T& collection)
{
while(collection.begin() != collection.end())
{
cout << "In the loop\n";
}

//T::iterator iter;
// ** Compiler error upon uncommenting **
}

//////////////////////////////////////////////////////
//
///////////////////////////////////////////////////////
main()
{
list<intcoll;
access_the_iterator(coll);
return 0;
}

On Fri, 05 Oct 2007 15:40:29 -0700, Generic Usenet Account wrote:

Quote:

Hi,
>
I am passing a container type (e.g. list<int>) to a function template
but I am getting a compiler error when I try to declare an iterator for
that type in the function. However, I am able to invoke begin() and
end() without any problem. Is there a way to declare an iterator for
the container type? Sample code follows.

[snip]

Quote:

template<typename T>
void
access_the_iterator(T& collection)
{
while(collection.begin() != collection.end()) {
cout << "In the loop\n";
}
>
//T::iterator iter;

Try

typename T::iterator iter;

Quote:

// ** Compiler error upon uncommenting **
}

--
Markus Schoder

Generic Usenet Account <usenet@sta.samsung.comwrote in
news:1191624029.029922.93220@57g2000hsv.googlegrou ps.com:

Quote:

Hi,
>
I am passing a container type (e.g. list<int>) to a function template
but I am getting a compiler error when I try to declare an iterator
for that type in the function. However, I am able to invoke begin()
and end() without any problem. Is there a way to declare an iterator
for the container type? Sample code follows.

That's what is referred to as a dependant type name. You need the
"typename" keyword:

typename T::iterator iter;

C++ needs the hint that T::iterator is a type and not a function or
variable name. (Remember, it doesn't know what T is when it parses that
line)

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STL::Accessing the iterator for a "passed" container type

re: STL::Accessing the iterator for a "passed" container type

re: STL::Accessing the iterator for a "passed" container type

re: STL::Accessing the iterator for a "passed" container type

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