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I'm not very good with namespaces, so my problem is probably a simple one.
I can't get the following to compile on gcc 3.3.1. It reports

main.cc: In function 'std:stream & operator<<(std:stream, const
ns::C&)':
main.cc:12: error: 'int ns::C::i' is private
main.cc:17: error: within this context
main.cc: In function 'int main(int, char**)':
main.cc:23: error: ambiguous overload for 'operator<<' in 'std::cout << c'
main.cc:16: error: candidates are: std:stream& operator<<(std:stream&,
const ns::C&)
main.cc:9: error: std:stream& ns:perator<<(std:stream&, const
ns::C&)
make: *** [main.o] Error 1

#include <iostream>

namespace ns {
class C {
public:
C(int i) : i(i) {};
~C() {};

friend std:stream &operator<<(std:stream
&stream, const C &c); // line 9
private:
int i; // line 12
};
}

std:stream &operator<<(std:stream &stream, const ns::C &c) {
stream << "C(" << c.i << ")" << std::endl; // line 17
return stream;
}

int main(int argc, char *argv[]) {
ns::C c(4);
std::cout << c; // line 23
return 0;
}

"Tim Partridge" <> wrote in message
news waterloo.ca...
> I'm not very good with namespaces, so my problem is probably a simple one.
> I can't get the following to compile on gcc 3.3.1. It reports
>
> main.cc: In function 'std:stream & operator<<(std:stream, const
> ns::C&)':
> main.cc:12: error: 'int ns::C::i' is private
> main.cc:17: error: within this context
> main.cc: In function 'int main(int, char**)':
> main.cc:23: error: ambiguous overload for 'operator<<' in 'std::cout << c'
> main.cc:16: error: candidates are: std:stream& operator<<(std:stream&,
> const ns::C&)
> main.cc:9: error: std:stream& ns:perator<<(std:stream&, const
> ns::C&)
> make: *** [main.o] Error 1
>
> #include <iostream>
>
> namespace ns {
> class C {
> public:
> C(int i) : i(i) {};
> ~C() {};
>
> friend std:stream &operator<<(std:stream
> &stream, const C &c); // line 9
> private:
> int i; // line 12
> };
> }
>
> std:stream &operator<<(std:stream &stream, const ns::C &c) {
> stream << "C(" << c.i << ")" << std::endl; // line 17
> return stream;
> }
>
> int main(int argc, char *argv[]) {
> ns::C c(4);
> std::cout << c; // line 23
> return 0;
> }

Put the implementation of the friend function in the same namespace...
something like...

#include <iostream>

namespace ns {
class C {
public:
C(int i) : i(i) {}

friend std:stream &operator << (std:stream &, const C &);
private:
int i;
};
}

namespace ns {
std:stream &operator << (std:stream &o, const C &c) {
o << "C (" << c.i << ")" << std::endl;
return o;
}
}

int main (int argc, char *argv[]) {
ns::C c (4);
std::cout << c;
}

// C (4)

Regards

Brian

"Tim Partridge" <> wrote in message
news waterloo.ca...
> I'm not very good with namespaces, so my problem is probably a simple one.
> I can't get the following to compile on gcc 3.3.1. It reports
>
> main.cc: In function 'std:stream & operator<<(std:stream, const
> ns::C&)':
> main.cc:12: error: 'int ns::C::i' is private
> main.cc:17: error: within this context
> main.cc: In function 'int main(int, char**)':
> main.cc:23: error: ambiguous overload for 'operator<<' in 'std::cout << c'
> main.cc:16: error: candidates are: std:stream& operator<<(std:stream&,
> const ns::C&)
> main.cc:9: error: std:stream& ns:perator<<(std:stream&, const
NB ns:perator<<
> ns::C&)
> make: *** [main.o] Error 1
>
> #include <iostream>
>
> namespace ns {
> class C {
> public:
> C(int i) : i(i) {};
> ~C() {};
>
> friend std:stream &operator<<(std:stream
> &stream, const C &c); // line 9

This says you have a friend called
ns:perator<<(std:stream&,const ns::C&)

> private:
> int i; // line 12
> };
> }
>
> std:stream &operator<<(std:stream &stream, const ns::C &c) {
> stream << "C(" << c.i << ")" << std::endl; // line 17
> return stream;
> }

This declares a (non-friend) global called
:perator<<(std:stream&,const ns::C&)

>
> int main(int argc, char *argv[]) {
> ns::C c(4);
> std::cout << c; // line 23

both :perator<< and ns:perator<< match hence ambiguity

> return 0;
> }

Personally I find it clearer never to use friends - declare a public print
method
and declare a non-friend operator<< to use it.