operator<< cannot use an IT member of its friend, array - c++

Question

i suppose that, if i make operator<< a friend of
a data struct (array by name);
//Forward Declarations
template<typename S, typename T>
struct array;
template<typename U, typename V>
ostream& operator<< (ostream& ous, const array<U, V>& t);
then, i will be able to do something like this; inside the implementation of operator <<
//operator<< is a friend of struct array{} already
template<typename T, typename U>
ostream& operator<< (ostream& os, const array<T, U>& var){
if(var){
/*Error: 'IT' was not declared in this scope*/
for(IT it = var.data.begin(); it != var.data.end(); it++){
/*and i thought i need not redeclare IT before using it
since operator<< is a friend of array already*/
}
}
else{cout << "empty";}
return os;
}
Now, here is array's implementation:
/*explicit (full) specialization of array for <type, char>*/
template<>
template<typename Key>
struct array<Key, char>{
//data members
map<const Key, string> data;
typedef map<const Key, string>::iterator IT;
//member function
friend ostream& operator<< <>(ostream& ous, const array& t);
//other stuff/functions
};
lastly, the compiler is angry when i test-drove it like this;
void Test(){
array<int, char> out;
out[1] = "one"; //Never mind. this has been taken care of
out[2] = "two";
cout << out; //Error: 'IT' was not declared in this scope
}
Question:
what exactly am i doing wrong, or, why can't i dirrectly access and use
IT (a typedef within array), even after i had declared operator << (that requests IT)
as a friend of the array struct ?

Answer 1

Write
for( typename array<T, U>::IT it = var.data.begin(); it != var.data.end(); it++){
And change
typedef map<const Key, string>::iterator IT;
to
typedef typename std::map<const Key, string>::const_iterator IT;
Here is a demonstrative program where instead of the std::map I used std::array for simplicity. I think it can help you.
#include <iostream>
#include <array>
template <typename T, size_t N>
struct A
{
std::array<T, N> a;
typedef typename std::array<T, N>::const_iterator IT;
};
template <typename T, size_t N>
std::ostream & operator <<( std::ostream &os, const A<T, N> &a )
{
for ( typename A<T, N>::IT it = a.a.begin(); it != a.a.end(); ++it ) os << *it << ' ';
return os;
}
int main()
{
A<int, 10> a = { { { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 } } };
std::cout << a << std::endl;
return 0;
}
The program output is
0 1 2 3 4 5 6 7 8 9

Answer 2

When you're using IT inside your template the compiler looks in the current scope (the operator template) for a declaration of a type named IT.
This fails because you defined the type as part of your array struct.
So in order to use the IT type you need to fully qualify it by using array<T,U>::IT.
Or you could try auto instead if you're using C++11.