Q:

templated function with an argument that is a templated class

I realise there are a lot of similar questions, but I couldn't find any that solves my problem (read: I did not truly understood the answers so that I could apply it to my case)
I have the following function:
void printVariable(map<string, bool*>::iterator it)
{
   cout << it->first << " " << *(it->second) << endl;
}

Now If i have a map<string, bool*> mapping I can do the following: printVariable(mapping.begin());
This works, now I also have a a map<string, int*> and want to be able to do the same, so I figured I change the printVariable function:
template <typename T>
void printVariable(map<string, T*>::iterator it)
{
   cout << it->first << " " << *(it->second) << endl;
}

However this gives compile errors (gcc):
error: variable or field 'printVariable' declared void
error: expected ')' before 'it'

I guess I can work around this pretty easy by overloading the function. But I'd like to know why the above is not working.
Edit2: removed text claiming a right solution was wrong

A:

You have to say typename to disambiguate the dependent name:
template <typename T>
void printVariable(typename map<string, T*>::iterator it)
//                 ^^^^^^^^

However, note that this is not a deducible context, so this form isn't terribly convenient.
Better yet, just make the entire iterator a template parameter:
template <typename Iter>
void printVariable(Iter it)
{ /* ... */ }

That way, you also capture maps with non-standard comparators or allocators, and unordered maps, etc. etc.

Here's a simple thought experiment for why you don't have a deducible context in the first situation: How should T be deduced in the following call of foo?
template <typename T> void foo(typename Bar<T>::type);

template <typename T> struct Bar { typedef char type; };
template <> struct Bar<int>      { typedef int type; };
template <> struct Bar<double>   { typedef int type; };

foo(10); // What is T?

