template specialization, nested?

[Schüle Daniel]

Hello all,

my question is basically wheather it's possible to
have nested templates .. I mean the following
if there are one templ. class and one global templ. function

is it possible to specialize the function for say <int>
but this for all possible classes (from class template)

the code is here

template <typename>
class X;

template <typename type>
void foo(X<type> &);

template <typename type = int>
class X
{
friend void foo<type>();
type x;
public:
X(type x) : x(x) {}
};

template <typename type>
void foo(X<type> & x)
{
cout << "with type = " << typeid(type).name() << endl;
cout << x.x << endl;
}

template <> // this seems not ok
template <typename type> // but i need "type" for X
void foo<void>(X<type> & x)
{
cout << "this is void overload" << endl;
cout << x.x << endl;
}

int main()
{
X<> x(1); // int
foo<void>(x);
return EXIT_SUCCESS;
}

Regards, Daniel

g++ --version
g++ (GCC) 4.0.2

[Victor Bazarov]

Schüle Daniel wrote:

my question is basically wheather it's possible to
have nested templates .. I mean the following
if there are one templ. class and one global templ. function

is it possible to specialize the function for say <int>
but this for all possible classes (from class template)
Uh... I am not sure I understand what you're trying to accomplish.

the code is here

template <typename>
class X;

template <typename type>
void foo(X<type> &);

template <typename type = int>
class X
{
friend void foo<type>();
There is no 'foo' without arguments! Did you mean

friend void foo<type>(X&);

?
type x;
public:
X(type x) : x(x) {}
};

template <typename type>
void foo(X<type> & x)
{
cout << "with type = " << typeid(type).name() << endl;
cout << x.x << endl;
}

template <> // this seems not ok
template <typename type> // but i need "type" for X
OK. So, you're trying to define a specialisation of 'foo' on 'void'
so that the argument is now a template. IOW, you're defining another
template here... That's not possible. Besides, there are no partial
specialisations of function templates.

Why do you think you need this construct?
void foo<void>(X<type> & x)
{
cout << "this is void overload" << endl;
cout << x.x << endl;
}

int main()
{
X<> x(1); // int
foo<void>(x);
return EXIT_SUCCESS;
}

Regards, Daniel

g++ --version
g++ (GCC) 4.0.2

V
--
Please remove capital As from my address when replying by mail

[Schüle Daniel]

Hello Victor,

template <typename>
class X;

template <typename type>
void foo(X<type> &);

template <typename type = int>
class X
{
friend void foo<type>();

There is no 'foo' without arguments! Did you mean

friend void foo<type>(X&);

yes, of course you are right

type x;
public:
X(type x) : x(x) {}
};

template <typename type>
void foo(X<type> & x)
{
cout << "with type = " << typeid(type).name() << endl;
cout << x.x << endl;
}

template <> // this seems not ok
template <typename type> // but i need "type" for X

OK. So, you're trying to define a specialisation of 'foo' on 'void'
so that the argument is now a template. IOW, you're defining another
template here... That's not possible. Besides, there are no partial
specialisations of function templates.

something like

template <typename type>
void foo() {}

template <>
void foo<int>(){}

my understanding of it is, that this is fully specialized
and is ok, my compiler at least doesn't complain
Why do you think you need this construct?

I thought it's time to get familar with template usage
(to write libraries)
so mostly this an academic case

void foo<void>(X<type> & x)

isn't it fully specialized?

Regards, Daniel

[Victor Bazarov]

Schüle Daniel wrote:

[..]
something like

template <typename type>
void foo() {}

template <>
void foo<int>(){}

my understanding of it is, that this is fully specialized
and is ok, my compiler at least doesn't complain

Yes. Fully specialised function templates are OK.

Why do you think you need this construct?

I thought it's time to get familar with template usage
(to write libraries)
so mostly this an academic case

void foo<void>(X<type> & x)

isn't it fully specialized?

How would it be if 'type' is unknown?

V
--
Please remove capital As from my address when replying by mail

[Schüle Daniel]

[..]

void foo<void>(X<type> & x)

isn't it fully specialized?

How would it be if 'type' is unknown?

my minds eye saw only
void foo<void>
:)

thanks for correction