Re: Stream operator in namespace masks global stream operator

From:

Fei Liu <feiliu@aepnetworks.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 09 May 2007 11:08:09 -0400

Message-ID:

<f1so4c$aks$1@aioe.org>

mrstephengross wrote:

Hi folks. I've got a weird situation--gcc doesn't like the folllowing
code snippet, but I don't know if it's correct or not. Here's the
situation:

In the global namespace, I've got a operator<< declared that will send
a vector<T> to a std::ostream.
In the "outer" namespace, I've got a operator<< declared that will
send a Thing<T> to a std::ostream.
In the "outer" namespace, I've got a function "foo" that tries to send
a vector<T> to a std::ostream.

When I try to compile it, gcc complains that there's no match for
operator<< in the foo function's definition.

Is this correct? Why is gcc not seeing the global namespace
operator<< ?

Thanks,
--Steve (sgross@sjm.com)

=== test.cpp ===

#include <iostream>
#include <vector>

template<typename T> std::ostream & operator<< (std::ostream & o,
const std::vector<T> & v);

namespace outer {

template<class T> class Thing { };

template<typename T> std::ostream & operator<< (std::ostream & o,
const Thing<T> & t);

void foo() { std::vector<double> v; std::cout << v; }

}

int main()
{
return 0;
}

=== EOF ===

The compiler is telling you it cannot find a prototype for '<<' in foo.
The '<<' function hides the '<<' function in global namespace. Try the
below implementation instead:

#include <iostream>
#include <vector>

template<typename T> std::ostream & operator<< (std::ostream & o,
const std::vector<T> & v);

namespace outer {

template<class T> class Thing { };

template<typename T> std::ostream & operator<< (std::ostream & o,
const T & t);

void foo() { std::vector<double> v; std::cout << v; }

}

int main()
{
return 0;
}