Strange template problem.

[Jason Heyes]

The following program does not compile. Apparantly "t" is inaccessible.

#include <iostream>
using namespace std;

template <class T> class Foo
{
T t;
public:
Foo(T t_) : t(t_) { }
void print() const { cout << t << endl; }
friend Foo<T> operator+(int n, const Foo<T> &f);
};

template <class T>
Foo<T> operator+(int n, const Foo<T> &f)
{
return Foo<T>(n + f.t);
}

int main()
{
Foo<int> f1(3), f2(5);
(1 + f2).print();
return 0;
}

This one does compile.

#include <iostream>
using namespace std;

template <class T> class Foo
{
public:
T t;
Foo(T t_) : t(t_) { }
void print() const { cout << t << endl; }
};

template <class T>
Foo<T> operator+(int n, const Foo<T> &f)
{
return Foo<T>(n + f.t);
}

int main()
{
Foo<int> f1(3), f2(5);
(1 + f2).print();
return 0;
}

And so does this one.

#include <iostream>
using namespace std;

template <class T> class Foo
{
T t;
public:
Foo(T t_) : t(t_) { }
void print() const { cout << t << endl; }
friend Foo<T> operator+(int n, const Foo<T> &f);
};

Foo<int> operator+(int n, const Foo<int> &f)
{
return Foo<int>(n + f.t);
}

int main()
{
Foo<int> f1(3), f2(5);
(1 + f2).print();
return 0;
}

Strange huh? Can someone please explain why the first program doesn't
compile? I can rewrite it into an equivalent version that does compile but I
am curious as to why that version does not. I am using VC++ 6.0. Thanks.

[tom_usenet]

On Mon, 15 Dec 2003 21:32:21 +1100, "Jason Heyes"
<ge******@optus net.com.au> wrote:

The following program does not compile. Apparantly "t" is inaccessible.

#include <iostream>
using namespace std;

template <class T> class Foo
{
T t;
public:
Foo(T t_) : t(t_) { }
void print() const { cout << t << endl; }
friend Foo<T> operator+(int n, const Foo<T> &f);
The above declares a non-template operator. I suspect this isn't what
you meant. See

http://gcc.gnu.org/faq.html#friend
Strange huh? Can someone please explain why the first program doesn't
compile? I can rewrite it into an equivalent version that does compile but I
am curious as to why that version does not. I am using VC++ 6.0. Thanks.

VC++6 has very poor template support, so the faq above might not help.
Try in an MSVC specific group for workarounds for that compiler. For a
quick suggestion, try defining the friend function inside the class
declaration.

Tom

C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html

[Dietmar Kuehl]

"Jason Heyes" <ge******@optus net.com.au> wrote:

The following program does not compile. Apparantly "t" is inaccessible.
Sure: it is inaccessible!
#include <iostream>
using namespace std;

template <class T> class Foo
{
T t;
't' is made private here...
public:
Foo(T t_) : t(t_) { }
void print() const { cout << t << endl; }
friend Foo<T> operator+(int n, const Foo<T> &f);
.... and this declares the non-template function 'operator+()' with certain
arguments to be a friend...
};

template <class T>
Foo<T> operator+(int n, const Foo<T> &f)
{
.... while the implementation is actually a template...
return Foo<T>(n + f.t);
.... and hence 't' is inaccessible.
}

You probably wanted to make the template 'operator+()' a friend. Since I'm
personally using friends only very rarely I'm a little bit rusty on what
is needed to make it legal but I think this should do:

template <class T> class Foo;
template <class T> Foo<T> operator+(int, Foo<T> const&);
template <class T> class Foo {
T t;
friend Foo<T> operator+ <T>(int, Foo<T> const&);
// ...
};

Quite painful but at least doable. However, it is a good argument to avoid
friends :-)
--
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