Question

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Where and why do I have to put “template” and “typename” on dependent names?

This is a specific instance of the question: Officially, what is typename for?

I am asking for the specific reason that the compiler is NOT aware that the following is a type:

#include <set>
#include <vector>

template<typename T>    // T is a type, right?
void f(const char name[], const std::vector<T>& foo) // typename NOT needed here
{
  for(std::set<T>::iterator itr =  // here, it is needed

If I declare:

std::set<int>::iterator itr; // no problem

The above clearly defines that T is a type, so why is typename required in one and not the other?

Answer 1

While you’ve links to a thorough answer, it’s a lot to wade through. So – put as simply as I can – the point is that the compiler does indeed know that T is a type as you say, but using that information and even having seen the included source code for std::set, it can’t be sure whether the iterator identifier inside the set<T> will name a type, a function, or a variable. This might seem surprising as if you look at the set<> template you can work it out, but remember that somewhere between the compiler parsing your f<>() template and before it’s instantiated, a specialisation for set<T> may be specified that uses identifier for a non-type, or simply lacks it altogether.

So, the typename keyword just tells the compiler, hey – whatever happens you can expect iterator to name a type, and perform some validatition of the f<>() template code on that basis without waiting to see an instantiation.