unnamed namespace problem

[Sandy]

Hi,
I have two files as folllows

file1.cpp
#include<iostream>
using namespace std;
namespace {
void show();
void fun() { cout<<"fun called\n"; }
}

int main()
{
show();
return 0;
}


file2.cpp
#include<iostream>
using namespace std;
namespace{
void fun();
void show(){
fun();
cout<<"show called\n";
}
}

While trying to run this the linker is giving following message
Undefined first referenced
symbol in file
(anonymous namespace)::fun() file2.o
(anonymous namespace)::show() file1.o
ld: fatal: Symbol referencing errors. No output written to a.out
collect2: ld returned 1 exit status


As far as i think, it should be able to find the definitions because
everything here belongs to a single "un-named" namespace.
Then why am i getting the problem?

 

[Greg]

Hi,
I have two files as folllows

file1.cpp
#include<iostream>
using namespace std;
namespace {
void show();
void fun() { cout<<"fun called\n"; }
}

int main()
{
show();
return 0;
}


file2.cpp
#include<iostream>
using namespace std;
namespace{
void fun();
void show(){
fun();
cout<<"show called\n";
}
}

While trying to run this the linker is giving following message
Undefined first referenced
symbol in file
(anonymous namespace)::fun() file2.o
(anonymous namespace)::show() file1.o
ld: fatal: Symbol referencing errors. No output written to a.out
collect2: ld returned 1 exit status


As far as i think, it should be able to find the definitions because
everything here belongs to a single "un-named" namespace.
Then why am i getting the problem?

Every unnamed namespace is unique - which means that every unnamed
namespace is a different namespace from every other unnamed namespace.
Not only across files and even across time. An unnamed namespace may
not even be the same unnamed namespace that it was before it was last
compiled.

In this case there are two show() functions and two fun() functions
declared, but fewer than four functions are actually defined.

Greg

 

[Ron Natalie]

Every unnamed namespace is unique - which means that every unnamed
namespace is a different namespace from every other unnamed namespace.
Not only across files and even across time. An unnamed namespace may
not even be the same unnamed namespace that it was before it was last
compiled.

That's NOT true.

All the unnamed namespaces in a single translation unit are the same
namespace. I'm not even sure what the point you're trying to make
about time is. Multiple unnamed namespaces in the same file are hence
the same namespace.

The namespace is however different from other namespaces in other
translation units (which is what the user has).

 

[Robbie Hatley]

Hi,
Hello.

I have two files as folllows

file1.cpp
#include<iostream>
using namespace std;

Ewwwwwwwww. That dumps thousands of names from namespace "std"
into the global namespace. Bad idea. (Creates danger of name
collision.) I wish textbook authors would stop telling people
to do that.

namespace {
void show();
void fun() { cout<<"fun called\n"; }
}

int main()
{
show();
return 0;
}

And the name of this namespace is??????????????
If you want to be able to access stuff in that
namespace from other files, you need to name it.
Try this instead:

// file1.cpp
#include<iostream>
using std::cout;
namespace MyNiftyNameSpace
{
void show();
void fun() {cout << "fun called\n";}
}

int main()
{
MyNiftyNameSpace::show();
return 0;
}

file2.cpp
#include<iostream>
using namespace std;
namespace{
void fun();
void show(){
fun();
cout<<"show called\n";
}
}

Once you've named a namespace, you can add to it in other
files, like this:

// file2.cpp
#include<iostream>
using std::cout;
namespace MyNiftyNameSpace
{
void fun();
void show(){
fun();
cout<<"show called\n";
}
}

While trying to run this the linker is giving following message
Undefined first referenced
symbol in file
(anonymous namespace)::fun() file2.o
(anonymous namespace)::show() file1.o
ld: fatal: Symbol referencing errors. No output written to a.out
collect2: ld returned 1 exit status


As far as i think, it should be able to find the definitions because
everything here belongs to a single "un-named" namespace.
Then why am i getting the problem?

You were using MULTIPLE un-named namespaces (one per file).
Interection equals null set. (Ie, the namespaces are disjoint.)

But if you name a namespace, you can add stuff to in in many files,
and the linker will hunt-down the missing pieces and link them
together for you.


Cheers,
Robbie Hatley
Tustin, CA, USA
email: lonewolfintj at pacbell dot net
web: home dot pacbell dot net slant earnur slant