Does the friend function have to be in the same file? - c++

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

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.

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

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;
}
}

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.

Related

Member is inaccessible

class Example{
public:
friend void Clone::f(Example);
Example(){
x = 10;
}
private:
int x;
};
class Clone{
public:
void f(Example ex){
std::cout << ex.x;
}
};
When I write f as a normal function, the program compiles successful. However, when I write f as a class member, this error occurs.
Screenshot:
The error you're seeing is not a root-cause compilation error. It is an artifact of a different problem. You're friending to a member function of a class the compiler has no earthly clue even exists yet,much less exists with that specific member.
A friend declaration of a non-member function has the advantage where it also acts as a prototype declaration. Such is not the case for a member function. The compiler must know that (a) the class exists, and (b) the member exists.
Compiling your original code (I use clang++ v3.6), the following errors are actually reported:
main.cpp:6:17: Use of undeclared identifier 'Clone'
main.cpp:17:25: 'x' is a private member of 'Example'
The former is a direct cause of the latter. But doing this instead:
#include <iostream>
#include <string>
class Example;
class Clone
{
public:
void f(Example);
};
class Example
{
public:
friend void Clone::f(Example);
Example()
{
x = 10;
}
private:
int x;
};
void Clone::f(Example ex)
{
std::cout << ex.x;
};
int main()
{
Clone c;
Example e;
c.f(e);
}
Output
10
This does the following:
Forward declares Example
Declares Clone, but does not implement Clone::f (yet)
Declares Example, thereby making x known to the compiler.
Friends Clone::f to Example
Implements Clone::f
At each stage we provide what the compiler needs to continue on.
Best of luck.

Is it possible to friend a class in an anonymous namespace in C++?

I am porting code from Java to c++ and I'd like to replicate some anonymous functionalities.
In file A.h I have :
class A
{
private:
int a;
class AnonClass;
friend class AnonClass;
};
In file A.cpp I have :
namespace
{
class AnonClass
{
public:
AnonClass(A* parent)
{
parent->a = 0; // This doesn't work, a is not accessible
}
}
}
Is it possible to friend a class in an anonymous namespace in C++?
In Java you can declare anonymous classes so it would be very similar. Also it would not expose AnonClass to clients of A.h
Less known alternative is to make class Anon a member class of A. Inside class A you only need a line class Anon; -- no real code, no friend declaration. Note it goes within class A, almost as in Java. In the .cpp file you write all the details about Anon but you put it not in anonymous namespace but withinA::
class A::Anon { ..... };
You can split declaration and implementation of A::Anon, as usual, just remeber always add A:: to Anon.
The class Anon is a member of A and as such gets access to all other members of A. Yet it remains unknown to clients of A and does not clutter global namespace.
As far as I can see you can not. The reasons:
The “anonymous” namespace is accessible only within the file you created it in.
You have to define whole AnonClass class and it's functions in one namespace, i.e. in one place in the program.
Class A has to be defined before AnonClass constructor.
AnonClass has to be at least declared before class A.
So you see, you can't break AnonClass definition on two parts. And you can't define it both before and after A class.
The only option - put class A into the same anonymous namespace. This code works:
namespace
{
class A
{
public:
A():a(0){};
private:
int a;
friend class AnonClass;
};
class AnonClass
{
public:
AnonClass(A* parent);
};
AnonClass::AnonClass(A* parent)
{
parent->a = 0;
};
}
int main() {
A a;
return 0;
}
I hope this helps.

Is pimpl compatible with anonymous namespaces?

I am trying to use the pimpl pattern and define the implementation class in an anonymous namespace. Is this possible in C++? My failed attempt is described below.
Is it possible to fix this without moving the implementation into a namespace with a name (or the global one)?
class MyCalculatorImplementation;
class MyCalculator
{
public:
MyCalculator();
int CalculateStuff(int);
private:
MyCalculatorImplementation* pimpl;
};
namespace // If i omit the namespace, everything is OK
{
class MyCalculatorImplementation
{
public:
int Calculate(int input)
{
// Insert some complicated calculation here
}
private:
int state[100];
};
}
// error C2872: 'MyCalculatorImplementation' : ambiguous symbol
MyCalculator::MyCalculator(): pimpl(new MyCalculatorImplementation)
{
}
int MyCalculator::CalculateStuff(int x)
{
return pimpl->Calculate(x);
}
No, the type must be at least declared before the pointer type can be used, and putting anonymous namespace in the header won't really work. But why would you want to do that, anyway? If you really really want to hide the implementation class, make it a private inner class, i.e.
// .hpp
struct Foo {
Foo();
// ...
private:
struct FooImpl;
boost::scoped_ptr<FooImpl> pimpl;
};
// .cpp
struct Foo::FooImpl {
FooImpl();
// ...
};
Foo::Foo() : pimpl(new FooImpl) { }
Yes. There is a work around for this. Declare the pointer in the header file as void*, then use a reinterpret cast inside your implementation file.
Note: Whether this is a desirable work-around is another question altogether. As is often said, I will leave that as an exercise for the reader.
See a sample implementation below:
class MyCalculator
{
public:
MyCalculator();
int CalculateStuff(int);
private:
void* pimpl;
};
namespace // If i omit the namespace, everything is OK
{
class MyCalculatorImplementation
{
public:
int Calculate(int input)
{
// Insert some complicated calculation here
}
private:
int state[100];
};
}
MyCalculator::MyCalculator(): pimpl(new MyCalculatorImplementation)
{
}
MyCalaculator::~MyCalaculator()
{
// don't forget to cast back for destruction!
delete reinterpret_cast<MyCalculatorImplementation*>(pimpl);
}
int MyCalculator::CalculateStuff(int x)
{
return reinterpret_cast<MyCalculatorImplementation*>(pimpl)->Calculate(x);
}
No, you can't do that. You have to forward-declare the Pimpl class:
class MyCalculatorImplementation;
and that declares the class. If you then put the definition into the unnamed namespace, you are creating another class (anonymous namespace)::MyCalculatorImplementation, which has nothing to do with ::MyCalculatorImplementation.
If this was any other namespace NS, you could amend the forward-declaration to include the namespace:
namespace NS {
class MyCalculatorImplementation;
}
but the unnamed namespace, being as magic as it is, will resolve to something else when that header is included into other translation units (you'd be declaring a new class whenever you include that header into another translation unit).
But use of the anonymous namespace is not needed here: the class declaration may be public, but the definition, being in the implementation file, is only visible to code in the implementation file.
If you actually want a forward declared class name in your header file and the implementation in an anonymous namespace in the module file, then make the declared class an interface:
// header
class MyCalculatorInterface;
class MyCalculator{
...
MyCalculatorInterface* pimpl;
};
//module
class MyCalculatorInterface{
public:
virtual int Calculate(int) = 0;
};
int MyCalculator::CalculateStuff(int x)
{
return pimpl->Calculate(x);
}
namespace {
class MyCalculatorImplementation: public MyCalculatorInterface {
...
};
}
// Only the ctor needs to know about MyCalculatorImplementation
// in order to make a new one.
MyCalculator::MyCalculator(): pimpl(new MyCalculatorImplementation)
{
}
markshiz and quamrana provided the inspiration for the solution below.
class Implementation, is intended to be declared in a global header file and serves as a void* for any pimpl application in your code base. It is not in an anonymous/unnamed namespace, but since it only has a destructor the namespace pollution remains acceptably limited.
class MyCalculatorImplementation derives from class Implementation. Because pimpl is declared as std::unique_ptr<Implementation> there is no need to mention MyCalculatorImplementation in any header file. So now MyCalculatorImplementation can be implemented in an anonymous/unnamed namespace.
The gain is that all member definitions in MyCalculatorImplementation are in the anonymous/unnamed namespace. The price you have to pay, is that you must convert Implementation to MyCalculatorImplementation. For that purpose a conversion function toImpl() is provided.
I was doubting whether to use a dynamic_cast or a static_cast for the conversion. I guess the dynamic_cast is the typical prescribed solution; but static_cast will work here as well and is possibly a little more performant.
#include <memory>
class Implementation
{
public:
virtual ~Implementation() = 0;
};
inline Implementation::~Implementation() = default;
class MyCalculator
{
public:
MyCalculator();
int CalculateStuff(int);
private:
std::unique_ptr<Implementation> pimpl;
};
namespace // Anonymous
{
class MyCalculatorImplementation
: public Implementation
{
public:
int Calculate(int input)
{
// Insert some complicated calculation here
}
private:
int state[100];
};
MyCalculatorImplementation& toImpl(Implementation& impl)
{
return dynamic_cast<MyCalculatorImplementation&>(impl);
}
}
// no error C2872 anymore
MyCalculator::MyCalculator() : pimpl(std::make_unique<MyCalculatorImplementation>() )
{
}
int MyCalculator::CalculateStuff(int x)
{
return toImpl(*pimpl).Calculate(x);
}

Local Classes in C++

I am reading "Local Classes" concept in Object-oriented programming with C++ By Balagurusamy (http://highered.mcgraw-hill.com/sites/0070593620/information_center_view0/).
The last line says "Enclosing function cannot access the private members of a local class. However, we can achieve this by declaring the enclosing function as a friend."
Now I am wondering how the highlighted part can be done?
Here is the code I was trying but no luck,
#include<iostream>
using namespace std;
class abc;
int pqr(abc t)
{
class abc
{
int x;
public:
int xyz()
{
return x=4;
}
friend int pqr(abc);
};
t.xyz();
return t.x;
}
int main()
{
abc t;
cout<<"Return "<<pqr(t)<<endl;
}
I know the code looks erroneous, any help would be appreciable.
Your friend statement is fine.
int pqr() {
class abc {
int x;
public:
abc() : x(4) { }
friend int pqr();
};
return abc().x;
}
int main() {
cout << "Return " << pqr() << endl;
}
Edit:
IBM offers this explanation for the issue raised in the comments:
If you declare a friend in a local class, and the friend's name is unqualified, the compiler will look for the name only within the innermost enclosing nonclass scope. [...] You do not have to do so with classes.
void a();
void f() {
class A {
// error: friend declaration 'void a()' in local class without prior decl...
friend void a();
};
}
friend void a(): This statement does not consider function a() declared in namespace scope. Since function a() has not been declared in the scope of f(), the compiler would not allow this statement.
Source: IBM - Friend scope (C++ only)
So, you're out of luck. Balagurusamy's tip only works for MSVC and similar compilers. You could try handing off execution to a static method inside your local class as a work-around:
int pqr() {
class abc {
int x;
public:
abc() : x(4) { }
static int pqr() {
return abc().x;
}
};
return abc::pqr();
}
There seems to be a misunderstand about local classes.
Normally there are here to help you within the function... and should NOT escape the function's scope.
Therefore, it is not possible for a function to take as an argument its own local class, the class simply isn't visible from the outside.
Also note that a variety of compilers do not (unfortunately) support these local classes as template parameters (gcc 3.4 for example), which actually prevents their use as predicates in STL algorithms.
Example of use:
int pqr()
{
class foo
{
friend int pqr();
int x;
foo(): x() {}
};
return foo().x;
}
I must admit though that I don't use this much, given the restricted scope I usually use struct instead of class, which means that I don't have to worry about friending ;)
I have no solution for the friend thing yet (don't even know if it can be done), but read this and this to find out some more about local classes. This will tell you that you cannot use local classes outside the function they are defined in (as #In silico points out in his answer.)
EDIT It doesn't seem possible, as this article explains:
The name of a function first introduced in a friend declaration is in the scope of the first nonclass scope that contains the enclosing class.
In other words, local classes can only befriend a function if it was declared within their enclosing function.
The friend int pqr(abc); declaration is fine. It doesn't work because the abc type has not been defined before you used it as a parameter type in the pqr() function. Define it before the function:
#include<iostream>
// By the way, "using namespace std" can cause ambiguities.
// See http://www.parashift.com/c++-faq-lite/coding-standards.html#faq-27.5
using namespace std;
// Class defined outside the pqr() function.
class abc
{
int x;
public:
int xyz()
{
return x=4;
}
friend int pqr(abc);
};
// At this point, the compiler knows what abc is.
int pqr(abc t)
{
t.xyz();
return t.x;
}
int main()
{
abc t;
cout<<"Return "<<pqr(t)<<endl;
}
I know you want to use a local class, but what you have set up will not work. Local classes is visible only inside the function it is defined in. If you want to use an instance of abc outside the pqr() function, you have to define the abc class outside the function.
However, if you know that the abc class will be used only within the pqr() function, then a local class can be used. But you do need to fix the friend declaration a little bit in this case.
#include<iostream>
// By the way, "using namespace std" can cause ambiguities.
// See http://www.parashift.com/c++-faq-lite/coding-standards.html#faq-27.5
using namespace std;
// pqr() function defined at global scope
int pqr()
{
// This class visible only within the pqr() function,
// because it is a local class.
class abc
{
int x;
public:
int xyz()
{
return x=4;
}
// Refer to the pqr() function defined at global scope
friend int ::pqr(); // <-- Note :: operator
} t;
t.xyz();
return t.x;
}
int main()
{
cout<<"Return "<<pqr()<<endl;
}
This compiles without warnings on Visual C++ (version 15.00.30729.01 of the compiler).

How to : Static variable in inherited class

class B {
public:
static int a;
};
class C:B {
};
I want to use a variable through any inherited classes but it has problem when I declare a.
B::B() {
a=1;
};
Do I do it right ?
Thanks for reading and waiting for your comments.
// I miss semicolons which is not the error I'm talking .
// This is an error when I try to delcare
class GameState {
public:
static int a = 1;
//...
};
Error 7 error C2864: 'CGameState::a' : only static const integral data members can be initialized within a class d:\my dropbox\work\#today\gdimario\gdimario\gamestate.h 18
I try to write a simple question which shows the problem I want instead of pasting my whole code.
You can use it directly like you did from both the derived and base class.
Perhaps your error is that you don't have semicolons at the end of your class declarations?
class B {
public:
static int a;
};
class C:B {
};
If you want to call it from an instance of C then you need to use public inheritance: (If nothing is specified private inheritance is assumed)
class C : public B {
};
To initialize a you need to do this (typically at the top of your corresponding .CPP file):
int B::a = 3;
You need to write in a CPP file:
int B::a;
And add the semicolons that Brad suggests. (Did you even compile your code? What did the compiler say?)
i think you ll get linker error.
since you have not defined of the static variable in the .cpp file.
e.g.
//hearer file
class X{
public : static int a ;
}
//impl file
int X::a(0);
....or...
For integral type you can also defined static variables when they are declared like:
class X{
public : static int a = 0;
}