Does the friend function have to be in the same file?

Question

I am actually testing a file and I have a situation, where I need to access some of the protected members of the class from main.cpp. I tried to add, main() as friend, didn't work out and learned that it wont work, so I moved everything in the main() to a test () and made the test() as friend. still it shows the error.

Example would be

 //--File.hpp

 namespace Files {

 class File {
          public:
                File(long word_):word(word_) {}
          protected:
                long word;
          private:
                friend int test();
 };
 }//ns:Files


 //--List_File.hpp

 namespace Files {
 class List_File :public File {
         public:
               List_File() : File(sizeof(int) + sizeof(long)) {}
         private:
              friend int test();
 };  
 }//ns:Files 



//--main.cpp

  using namespace Files;

  int test() {

       File *pd = new List_File();
       assert(pd->word == 12); //LINE 34
       return 0;
  }

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

//it says error on Line 34: Base::value is protected. Please advice.

    g++ -O -Wall -Wno-unused -o a.out File.cpp List_File.cpp Data_File.cpp               
    Free_List_File.cpp main.cpp
    File.hpp: In function ‘int test()’:
    File.hpp:30:7: error: ‘long int Files::File::word’ is protected
    main.cpp:34:16: error: within this context
    make: *** [a.out] Error 1

Answer 1

No, it definitely doesn't have to be in the same file, but it obviously has to "know" what the class is, i.e.: the header that has the class definition should be included in the file where the function is implemented. Your code should be fine, as commented.

after you added some context

Your test function is not in the Files namespace. If you want it to be in the global context, you should treat it as "::test" within the namespace, otherwise the compiler might expect the "Files::test" to be the friend function, and not the "::test" as in your case. Can't find the formal standard reference, but I'm pretty sure that's the case. Note that you're performing a forward declaration here, so there's no default fall-back to the upper level of scope for name resolution.

Answer 2

Maybe you missing inheritance declaration in Derived class?

 class Derived : public Base {

I've tested your code (with inheritance declaration included) and it produced no error

Answer 3

littedev is right!! Updated the code according to the comments by littedev..

//--File.hpp   
namespace Files {   
    class File {
        public:
            File(long word_):word(word_) {}
        protected:
            long word;
        private:
            friend int test();
    };  
}//ns:Files    

//--List_File.hpp   

namespace Files {
    class List_File :public File {
        public:
            List_File() : File(sizeof(int) + sizeof(long)) {}          \
        private:
            friend int test();  
    };    
}//ns:Files     

//--main.cpp    

namespace Files{    
    int test() {         
        File *pd = new List_File();
        assert(pd->word == 12); //LINE 34
        return 0;
    }

    int main() {
        Files::test();
        return 0;
    } 
}

Answer 4

I would guess that you are trying to access a protected variable outside of the scope of the Files class definition. I would recommend a function that returns the variable word when it is called and use that instead of trying to access a protected variable outside of a class definition. I could be wrong in that I am not really sure what is the scope of a protected variable (whether it is limited only to class declarations or whether it can be accessed outside of the class definition) but I am pretty sure that I am right because protected variables are like private variables. They are only accessible within the class scope. Correct me if I am wrong.

EDIT: Oh I am sorry I didn't realize what you were doing. littleadv is right, your function declaration isn't within the files namespace.