friend function in templates

Forget it...
Board search gave the answer...
should have read the search result completely
When i try outside class defenition (as below) I get errors.
How can I solve it?

Code:

//Sample prog which generates same errors as in the original
#include <iostream>
using namespace std;

template<int n>
class A{
    private:
    int array[n];
    public:
    A(){
        for(int i=0; i<n; ++i)
            array[i] = i;
    }
    friend ostream& operator<<(ostream&, const A&);
};

template<int n>
ostream& operator<<(ostream& out, const A<n>& a){   //What is Wrong Here???
    for(int i=0; i<n; ++i)
        out<<a.array[i]<<",\t";
    return out;
}

int main(){
A<10> a1;
cout<<a1;
return 0;
}
main.cpp:13: warning: friend declaration `std::ostream& 
   operator<<(std::ostream&, const A<n>&)' declares a non-template function.

main.cpp:13: warning: (if this is not what you intended, make sure the function 
   template has already been declared and add <> after the function name here) 
//what does this mean???
   -Wno-non-template-friend disables this warning.

main.cpp: undefined reference to `operator<<(std::ostream&, A<10> const&)'

When I try inside class defenition It works all right.

Code:

#include <iostream>
using namespace std;

template<int n>
class A{
    private:
    int array[n];
    public:
    A(){
        for(int i=0; i<n; ++i)
            array[i] = i;
    }
    friend ostream& operator<<(ostream& out, const A<n>& a){
        for(int i=0; i<n; ++i)
            out<<a.array[i]<<"\t";
        return out;
    }
};

int main(){
A<10> a1;
cout<<a1;
return 0;
}

I refered some books but none of them have examples on outside class defenitions of templated friend functions
__________________________________________
I did a board search and I found this:
overloading the output operator <<
and also this in the same post:

using namespace std;

template <typename T, int Num = 10>
class Bar
{
T x_;
public:
Bar(T i) : x_(i) {}
const T& x() const { return x_; }
};

template <typename T, int Num>
ostream& operator<<(ostream& o, const Bar<T,Num> &b)
{
return o<<"Num = "<<Num<<" X = "<<b.x()<<endl;
}

how come operator<< is'nt a 'friend'?
x is accessed indirectly...