Define friend operator << for class template.

[Joe Hesse]

Hi,

I have a template class and I want to define an operator << as a friend
function. For each instantiation of the class I want a corresponding
instantiation of operator <<.
The following example fails to compile with g++ version 4.1.2.
I would appreciate it if you could help me fix it or point me to a suitable
reference.

Thank you,
Joe Hesse

*************** *************** **************
#include <iostream>

// forward declaration
template <typename T>
class Foo;

// forward declaration
template <typename T>
std::ostream & operator << (std::ostream &, const Foo<T&);

template <typename T>
class Foo {
private:
T value;
public:
Foo(const T & v) : value(v) {}
friend std::ostream & operator << <(std::ostrea m &, const Foo<T&);
};

// implement operator <<
template <typename T>
std::ostream & operator << (std::ostream &o, const Foo<T&f) {
return o << f.value ;
}

int main() {
Foo<intfi;
std::cout << fi;

return 0;
}

/* Here are the compiler error messages
Test.cpp: In function int main():
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(con st T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(c onst Foo<int>&)
*/

*************** *************** **************

[Joe Hesse]

Please forgive me, I have an obvious error. No response is needed from the
newsgroup.

"Joe Hesse" <jo*******@actc x.comwrote in message
news:13******** *****@corp.supe rnews.com...

Hi,

I have a template class and I want to define an operator << as a friend
function. For each instantiation of the class I want a corresponding
instantiation of operator <<.
The following example fails to compile with g++ version 4.1.2.
I would appreciate it if you could help me fix it or point me to a
suitable reference.

Thank you,
Joe Hesse

*************** *************** **************
#include <iostream>

// forward declaration
template <typename T>
class Foo;

// forward declaration
template <typename T>
std::ostream & operator << (std::ostream &, const Foo<T&);

template <typename T>
class Foo {
private:
T value;
public:
Foo(const T & v) : value(v) {}
friend std::ostream & operator << <(std::ostrea m &, const Foo<T&);
};

// implement operator <<
template <typename T>
std::ostream & operator << (std::ostream &o, const Foo<T&f) {
return o << f.value ;
}

int main() {
Foo<intfi;
std::cout << fi;

return 0;
}

/* Here are the compiler error messages
Test.cpp: In function int main():
Test.cpp:26: error: no matching function for call to Foo<int>::Foo()
Test.cpp:16: note: candidates are: Foo<T>::Foo(con st T&) [with T = int]
Test.cpp:12: note: Foo<int>::Foo(c onst Foo<int>&)
*/

*************** *************** **************

[want.to.be.professer]

Once you not provide a constructor ,There will be a default
constructor;
When you provide, There will be your provided constructor only;