Re: why is operator<< not found

I've tried the below code with MSVC and Comeau online compiler.
Both complain that operator<< for Outer<part<size_t> >::inner is not
defined. So how do I declare it without doing full specialization?

I've tried a template on any type, like Outer<AnyT>::inner (see
below) Also tried a template on the templated type: Outer<part<X>
 >::inner
// (see below) I can't find a typo? What is the rule in operation here?

Thanks,
homsan
---------------- 8< -----------------

#include <iostream>

template<class SizeT>
struct part {
typedef SizeT size_type;
size_type size;
};

template<class PartT>
class Outer
{
public:
typedef PartT part_type;
struct inner {
part_type p;
};
inner get() const { return inner(); }
};

// WHY is this not found in main()?

template<class PartT>
inline std::ostream& operator<<(std::ostream& ostr, typename
Outer<PartT>::inner const& it) {
ostr << it.p.size << " by Jove\n";
return ostr;
}

// Nor is this found - what gives?

template<class SizeT>
inline std::ostream& operator<<(std::ostream& ostr, typename
Outer<part<SizeT> >::inner const& it) {
ostr << it.p.size << " by Jove\n";
return ostr;
}

// Now this works, but of course I don't want to redefine for every
specialization... /*
std::ostream& operator<<(std::ostream& ostr, Outer<part<size_t>


ostr << it.p.size << " by Jove\n";
return ostr;
}
*/

int main() {
    Outer<part<size_t> > yippie;
    std::cout << yippie.get();
    return 0;
}