Thread: about typename keyword

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.

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'.