名称空间，类和自由函数 - 何时需要完全限定名称

Question

In my example below, why do I have to fully qualify the name of the free function in the cpp to avoid linker errors and why does it work for the class function without? Can you explain the difference?

ctest.h:

namespace Test
{
    int FreeFunction();

    class CTest
    {
        public:
            CTest();
            ~CTest();
    };
}

ctest.cpp:

#include "ctest.h"

using namespace Test;

// int FreeFunction()     -> undefined reference error
int Test::FreeFunction()  -> works just fine
{
    return 0;
}

CTest::CTest()                -> no need to fully qualify name, i.e. Test::CTest
{}

CTest::~CTest()
{}

Thanks for your time & help.

Answer 1

int FreeFunction(void);  

is just a declaration whereas the below is a definition.

class CTest 
{ 
    public: 
        CTest(); 
        ~CTest(); 
}; 

If you want to provide definition for an already declared entity in a namespace (eg in an enclosing namespace), it has to be fully qualified name.

EDIT2:

Here is something that would give you some more clarity. Note no using directive in this code.

namespace Test { 
    int FreeFunction(void);   // declare

    class CTest;              // declare
} 

int Test::FreeFunction(){return 0;} // define
class Test::CTest{            // define
};

int main(){}

EDIT 3: Declaration vs Definition (C++0x) $3.1/2-

A declaration is a definition unless it declares a function without specifying the function's body (8.4) , it contains the extern specifier (7.1.1) or a linkage-specification25 (7.5) and neither an initializer nor a function-body, it declares a static data member in a class definition (9.4), it is a class name declaration (9.1) , it is an opaque-enum-declaration (7.2), or it is a typedef declaration (7.1.3), a using-declaration (7.3.3), a static_assert-declaration (Clause 7), an attribute-declaration (Clause 7), an empty-declaration (Clause 7), or a using-directive (7.3.4).

Answer 2

While FreeFunction will resolve to Test::FreeFunction if you refer to it or call it after providing the using namespace Test; line, as far as defining the function goes, the compiler has no way to know if you're defining an entirely new function FreeFunction outside of any namespace, or whether you're defining the already declared Test::FreeFunction . The compiler defaults to thinking that you're defining an entirely new function.

For CTest::CTest , however, you're already referring to the class Test::CTest , and since there's no class or namespace CTest outside of the Test namespace, well, the reference to CTest::anything is unambiguous. So it knows that the constructor and destructor definitions refer to the in-namespace class CTest .

I think it's a small price to pay, to have to write Test::FreeFunction .

Hope this helps!

Answer 3

If you don't qualify FreeFunction definition, the compiler does not know for sure anther you want to provide implementation for the previously forward-declared Test::FreeFunction or for a separate FreeFunction in the current namespace.

On the other hand, there's only one way to resolve the name CTest - as the class definition from the Test namespace. Thus, there's no need to fully qualify it.

However, if the CTest name resolution is ambiguous (say there's another CTest class in the current namespace as well), you will have to fully qualify the method declarations.

Answer 4

When implementing a function it is usually preferable, I find, to open the namespace. Remember you can re-open them...

// in Test.cpp
namespace Test
{
   int FreeFunction()
   {
       return 0;
   }
}