Should this compile and link?

[russ]

Been looking at some code.....

#include <vector>
#include <iostream>
#include <algorithm>

template <typename T> struct A
{
struct B
{
std::vector< T > innervec;
} ;

std::vector< B > outervec;
};

template < typename T >
std::ostream & operator<< ( std::ostream & os, const typename A<T>::B &
b )
{
std::copy( b.innervec.begin(), b.innervec.end(),
std::ostream_iterator<T>( os, "\t" ) );
return os;
}

template < typename T >
std::ostream & operator<< ( std::ostream & os, const A<T>& a )
{
typedef typename A<T>::B B_t;

std::copy( a.outervec.begin(), a.outervec.end(),
std::ostream_iterator< B_t >( os, "\n" ) );
return os;
}

int main()
{
A<int> a;
A<int>::B b1;
A<int>::B b2;
b1.innervec.push_back( 11 );
b1.innervec.push_back( 12 );
b2.innervec.push_back( 21 );
b2.innervec.push_back( 22 );
a.outervec.push_back( b1 );
a.outervec.push_back( b2 );
std::cout << a;
}

Is that standard conforming?
Should it work fine as is?
My copy of MSVC 7.0 doesn't complain but Comeau's online beta says no
and fails when -tused is used.
If this is not supposed to work then why is that and what is the best
way of rewriting it?

[Victor Bazarov]

ru**@haydon449.plus.com wrote:

Been looking at some code.....
[...]
template < typename T >
std::ostream & operator<< ( std::ostream & os, const typename A<T>::B
'const typename A<T>::B &' is not one of the forms which the Standard
allows for template argument deduction from a function argument. So,
the compiler simply fails to deduce 'T' here, most likely.

& b )
[...]

Is that standard conforming?
I don't think so.
Should it work fine as is?
I don't think so.
My copy of MSVC 7.0 doesn't complain but Comeau's online beta says no
and fails when -tused is used.
If this is not supposed to work then why is that and what is the best
way of rewriting it?

Why should 'B' be inner to A<T>? Perhaps you should make 'B' a template
as well? The only work-around I can think of is to make 'B' a class not
nested in 'A<T>' but a separate class.

V