Re: question about stream operator and namespaces

From:
Barry <dhb2000@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sun, 23 Dec 2007 14:14:47 CST
Message-ID:
<fkkm17$ln9$1@news.cn99.com>
Russell McManus wrote:

I have a class with a stream operator which works as expected when the
class definition is not defined within a namespace. However, it does
not work as expect when defined within a namespace.

I am using gcc 3.4.6.

I include a short example program that shows the issue. On my system,
when the below program is saved as foo.cpp, the following compilation
line works and the program runs:

  g++ -DUSE_NAMESPACE=0 -o foo foo.cc && ./foo

However the following compilation line fails:

  g++ -DUSE_NAMESPACE=1 -o foo foo.cc && ./foo

Any ideas what I'm doing wrong?

-russ

Here is program:

#include <iostream>
#include <iomanip>
#include <sstream>
#include <iterator>
#include <algorithm>
#include <inttypes.h>
#include <vector>

using namespace std;

#ifndef USE_NAMESPACE
# define USE_NAMESPACE 0
#endif

#if (USE_NAMESPACE)

namespace foo
{
  template <class T>
  class Wrapper
  {
  public:
    Wrapper() {}
    explicit Wrapper(T const& t) : _t(t) {}
    T const& operator()() const { return _t; }
  private:
    T _t;
  };
}

template <class T>
ostream& operator<<(ostream& os, foo::Wrapper<T> const& t) {
  os << t(); return os;
}
typedef foo::Wrapper<uint32_t> Seqno;

#else

template <class T>
class Wrapper
{
public:
  Wrapper() {}
  explicit Wrapper(T const& t) : _t(t) {}
  T const& operator()() const { return _t; }
private:
  T _t;
};

template <class T>
ostream& operator<<(ostream& os, Wrapper<T> const& t) {
  os << t(); return os;
}
typedef Wrapper<uint32_t> Seqno;

#endif

int main(int argc, char* argv[]) {
  vector<Seqno> seqno;
  seqno.push_back(Seqno(1));
  seqno.push_back(Seqno(2));
  copy(seqno.begin(), seqno.end(), ostream_iterator<Seqno>(cout, "\n"));
}


ostream& operator<<(ostream& os, foo::Wrapper<T> const& t);
you have qualified "foo::wrapper<T>" and having std namespace using
derective, so ADL is not applied here.

two solutions:

1) make "std::ostream" unqualified, that's to move this "operator<<"
into namespace std

namespace std {
   ostream& operator<<(ostream& os, foo::Wrapper<T> const& t) {
     ...
   }
}

2) make "foo::Wrapper<T>" unqualified, that's to move this "operator<<"
into namespace foo

namespace foo {
   std::ostream& operator<<(std::ostream& os, Wrapper<T> const& t) {
     ...
   }
}

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