Doubt about namespaces

[Carlos Martinez]

//general.h
#ifndef _GENERAL_H
#define _GENERAL_H

namespace ip {

bool esServicioIP();

}

#endif //_GENERAL_H

//general.cc
#include <general.h>

....

using namespace ip;

bool esServicioIP() {
....
}

//distsgip.cc

.....

if(esServicioIP ()) {
...
}


Code compiles ok, but linker gives me an error saying ip::esServicioI P
is undefined.
Instead if I put namespace ip when defining esServicioIP:

bool ip::esServicioI P() {
....
}
Code compiles and links ok.

In many other pieces of code with clases instead of functions inside a
namespace, I don't need to qualify class' methods with namespace:

bool MyClass::myMeth od() {
....
}

Compiles and links ok

Is there any rule (perhaps of naming resolution or something else)
different for classes and normal functions?

Thanks in advance

[David Harmon]

Re: Doubt about namespaces

On Tue, 12 Sep 2006 10:13:43 +0200 in comp.lang.c++, Carlos Martinez
<cmg250@nospam. tid.eswrote,

>using namespace ip;
>
>bool esServicioIP() {
>...
>}

That function is not "inside" the ip namespace. It is a new
definition of its own independent function.

>In many other pieces of code with clases instead of functions inside a
>namespace, I don't need to qualify class' methods with namespace:
>
>bool MyClass::myMeth od() {
>...
>}

That function must be a member of some namespace or class known as
MyClass that already exists. It's not a valid declaration of
MyClass. Furthermore, MyClass::myMeth od must have already been
declared inside the namespace or class.

>Is there any rule (perhaps of naming resolution or something else)
>different for classes and normal functions?

Not really. For example,

using namespace ip;
MyClass foo; // has no connection with any ip::foo
class bar; // has no connection with any ip::bar

[Carlos Martinez]

Re: Doubt about namespaces

David Harmon wrote:

On Tue, 12 Sep 2006 10:13:43 +0200 in comp.lang.c++, Carlos Martinez
<cmg250@nospam. tid.eswrote,
>

>using namespace ip;
>>
>bool esServicioIP() {
>...
>}

>
That function is not "inside" the ip namespace. It is a new
definition of its own independent function.

Excuse me. I have ommited code in .h because I have considered obvious.

In .h

namespace ip {

bool esServicioIP();

}

>

>In many other pieces of code with clases instead of functions inside a
>namespace, I don't need to qualify class' methods with namespace:
>>
>bool MyClass::myMeth od() {
>...
>}

In .h

namespace ip {

class MyClass {
bool myMethod();
}

}

>
That function must be a member of some namespace or class known as
MyClass that already exists. It's not a valid declaration of
MyClass. Furthermore, MyClass::myMeth od must have already been
declared inside the namespace or class.
>

>Is there any rule (perhaps of naming resolution or something else)
>different for classes and normal functions?

>
Not really. For example,
>
using namespace ip;
MyClass foo; // has no connection with any ip::foo
class bar; // has no connection with any ip::bar
>

[David Harmon]

Re: Doubt about namespaces

On Tue, 12 Sep 2006 13:36:31 +0200 in comp.lang.c++, Carlos Martinez
<cmg250@nospam. tid.eswrote,

>David Harmon wrote:

>On Tue, 12 Sep 2006 10:13:43 +0200 in comp.lang.c++, Carlos Martinez
><cmg250@nospam .tid.eswrote,
>>

>>using namespace ip;
>>>
>>bool esServicioIP() {
>>...
>>}

>>
>That function is not "inside" the ip namespace. It is a new
>definition of its own independent function.

>
>Excuse me. I have ommited code in .h because I have considered obvious.
>
>In .h
>
>namespace ip {
>
bool esServicioIP();
>
>}

You are quite right, it is obvious enough. You have two functions.

In .h you declare ip::esServicioI P().
in .cc you define esServicioIP().

There is no reason you cannot have both of them. The compiler can
only hope you define ip::esServicioI P() somewhere too.

[F.J.K.]

Re: Doubt about namespaces

Carlos Martinez wrote:

Is there any rule (perhaps of naming resolution or something else)
different for classes and normal functions?
>
Thanks in advance

No. Try harder to get a minimum testcase, as required by the FAQ and
you'll see.