about typename keyword

I am reading thinking in c++ book(Vol.2).I am in template section amd i don't understand typename keyword's meaning and usage.

Here is sample code:

Code:

---

#include <iostream>
#include <list>
#include <memory>
#include <vector>
using namespace std;
 
template<class T, template<class U, class = allocator<U> >
        class Seq>
void printSeq(Seq<T>& seq) {
  for(typename Seq<T>::iterator b = seq.begin();
      b != seq.end();)
    cout << *b++ << endl;
}
 
int main() {
  // Process a vector
  vector<int> v;
  v.push_back(1);
  v.push_back(2);
  printSeq(v);
  // Process a list
  list<int> lst;
  lst.push_back(3);
  lst.push_back(4);
  printSeq(lst);
} ///:~

---

Here why do we write "typename Seq<T>::iterator b" .Maybe i don't understand because Seq<T> is special library.Is it a type of library?

I really don not understand typename keyword.

Can someone explain me that please?
Thanks.

In general it is not possible for the non-psychic compiler to know that Seq<T>::iterator is intended as the name of a type (type name) rather than the name of something that is not a type (such as a function or a template). If we want to state that something should be treated as a type, we can do so using the typename keyword. For more details you can read a C++ standard text. I got the above information from The C++ Programming Language, Special Edition by Stroustrup, the Creator of C++, section C.13.5 Typename and Template.

Quote:

---

I am in template section amd i don't understand typename keyword's meaning and usage.

---

Suppose you had a static member of the class Seq<T> called 'iterator' and it was equal to 10. You could display iterator's value like this:

cout<<Seq<T>::iterator;


So in this statement:

Seq<T>::iterator b = seq.begin();

the compiler thinks Seq<T>::iterator is a variable name, and it tries to replace it with its value. If there were a static variable in the class Seq<T> called iterator, and it had a value of 10, that would produce:

10 b = seq.begin();

Even though that doesn't make any sense, the compiler still can't figure out that in that context Seq<T>::iterator should not be considered a static variable name. So to tell the compiler that you are not refering to a static variable name, you need to use the keyword 'typename'. The typename keyword tells the compiler to look in Seq<T> for a type called 'iterator' rather than a variable called 'iterator'.

Ok friends ,now i understand.
Thanks fischerandom.
7stud it is very good description, thank you.