Friendship and Interfaces - c++

I'm having a compilation error using Apple's Clang 7.0 with the following friendship code, using C++11 standard. I wonder what's really wrong with it since it seems to be valid for me. I'm describing the setting and the error I'm having:
MyInterface
namespace namespace1
{
class MyInterface
{
friend class MyClass;
public:
virtual void some_method(void) = 0;
...
private:
type some_attribute;
...
}
}
MyClass::MyMethod Implementation
namespace namespace2
{
void MyClass::MyMethod(MyInterface* MyConcrete)
{
...
// MyConcrete implements MyInterface
if(MyConcrete->some_attribute == some_value) // Error*
{
...
}
...
}
}
Error
error: 'some_attribute' is a private member of 'namespace1::MyInterface'
I really expected that MyClass would have access to some_attribute in MyConcrete (which implemented MyInterface) regardless of the class access modifier. Any clues why this error is happening? Any suggestions?
Thank you!

MyClass is in namespace2. So you need to use:
friend class namespace2::MyClass;
You may also need to use forward decleration of MyClass before you define MyInterface.
Here's an example that compiles:
// forward decleration
namespace namespace2
{
class MyClass;
}
namespace namespace1
{
class MyInterface
{
friend class namespace2::MyClass; // added missing namespace
public:
virtual void some_method(void) = 0;
private:
int some_attribute;
};
}
namespace namespace2
{
class MyClass
{
void MyMethod(namespace1::MyInterface* MyConcrete)
{
if(MyConcrete->some_attribute == 1)
{
}
}
};
}
int main()
{
}
You can run it here.

friend class MyClass; in the context of ::namespace1::MyInterface refers to the class ::namespace1::MyClass, which is a different class from ::namespace2::MyClass. (That's the whole point of namespaces, right?)
Change the friend declaration to read like this:
friend class ::namespace2::MyClass;
Note that this requires that the type ::namespace2::MyClass has already been declared, so you either need to forward-declare it (namespace namespace2 { class MyClass; }) or you need to make sure that the definition is included prior to the definition of ::namespace1::MyInterface.
(See this demo.)

Related

how to order the definitions and declaration of class and friend function?

I have a function that is friend of a class. Therefore the function can access all class member.
However, I am confused with the order of definition.
If the function comes before the class then how do could it even get the class member? if the class come before the function, then it cannot know the function.
How to solve such cross reference conflict?
So I have code
class BP_Interface
{
friend void CommonUtility::HandleToFrame(Handle&);
private:
Handle frame_in_model;
}
// followed by definition
namespace CommonUtility
{
void HandleToFrame(Handle& eR)
{
BP_Interface().frame_in_model = eR;
return;
}
}
visual studio IDE shows error, and says frame_in_model cannot be accessed.
Declare CommonUtility::HandleToGraph before the definition of BP_Interface and define it later.
namespace CommonUtility
{
void HandleToGraph(Handle& eR);
}
class BP_Interface
{
friend void CommonUtility::HandleToFrame(Handle&);
private:
Handle frame_in_model;
};
// followed by definition
namespace CommonUtility
{
void HandleToGraph(Handle& eR)
{
BP_Interface().frame_in_model = eR;
return;
}
}

Segfault trying to work with shared_ptr in forward declared class

I'm having an odd issue where when I forward declare a class and run it, it calls the constructor successfully but when it tries to work with a shared_ptr, it segfaults within the __exchange_and_add on return __atomic_fetch_add forward declared class constructor.
The class we're trying to use is located in a linked shared object.
For example:
class MyForwardDeclaredClass {
public:
MyForwardDeclaredClass();
}
namespace MyNamespace
{
class MyClass
{
public:
MyClass() { _myForwardDeclaredClass.reset(new MyForwardDeclaredClass()); }
std::shared_ptr<MyForwardDeclaredClass> _myForwardDeclaredClass;
}
}
But if I do this:
#include <MyForwardDeclaredClass.hpp>
namespace MyNamespace
{
class MyClass
{
public:
MyClass() { _myForwardDeclaredClass.reset(new MyForwardDeclaredClass()); }
std::shared_ptr<MyForwardDeclaredClass> _myForwardDeclaredClass;
}
}
It works properly.

Friending/Using a class in a different namespace

If the class G is in the namespace GSpace and it needs to be friends with the class M in the global namespace what do you have to do? I thought this would work:
/////////////////////M.h//////////////////////
#include "GFile.h"
class M
{
public:
friend class GSpace::G; // doesn't work, throws error
}
After researching on StackOverflow a bit, I found this answer https://stackoverflow.com/a/3843743/1797424
/////////////////////M.h//////////////////////
namespace GSpace
{
class G;
}
class M
{
public:
friend class GSpace::G; // works, no error
private:
GSpace::G gClassMember; // errors: "M uses undefined class GSpace::G"
};
// Note that G.h includes M.h so I do not include it here,
// instead I have it included in M.cpp
That does work for getting the class friended. However, it creates an issue when I actually declare a class member using that type, because the class is not defined. GFile.h
Am I misunderstanding how #include and forward declaration behaves, or is it some kind of implementation issue on the side of the compiler (not likely, I'm assuming)?
Because your member is not a pointer or reference, the compiler needs to know the size of G. You can't use a forward declaration.
As noted in the comment, you need to qualify G with the namespace.
Here is code which compiles for me:
namespace GSpace
{
class G
{
};
}
class M
{
public:
friend class GSpace::G;
private:
GSpace::G gClassMember;
};
int main() {return 0;}
Try the following
namespace GSpace
{
class G;
}
class M
{
public:
friend class GSpace::G;
}
namespace GSpace
{
class G { /* definition of the class */ };
}

want friend function in namespace, but not the whole class

this is a follow-up question to friend function within a namespace
If I want testFunc to be in namespace TestNamespace, but I don't want TestClass also in TestNamespace, how can I declare them?
1.This doesn't work ("expected unqualified-id before namespace" on line 2)
class TestClass {
namespace TestNamespace {
friend void testFunc(TestClass &myObj);
};
}
namespace TestNamespace {
void testFunc(TestClass &myObj);
}
2.But it complains that "TestNamespace::testFunc(...) should have been declared inside 'TestNamespace'" when I do this-
class TestClass {
friend void TestNamespace::testFunc(TestClass &myObj);
}
namespace TestNamespace {
void testFunc(TestClass &myObj);
}
3.The above code is okay if I don't have the "TestNamespace::" in the friend function declaration on line 2, but I'm not sure if it then knows that TestNamespace::testFunc is a friend.
I don't know what's happening in any of these cases. Any insight would be much appreciated too. Thanks!
Try using forward declarations:
class TestClass;
namespace TestNamespace {
void testFunc(TestClass &myObj);
}
class TestClass {
friend void TestNamespace::testFunc(TestClass &myObj);
};
I don't know what "leda" was in your original example, but hopefully you can substitute whatever names you need given the above.
You can make it simple:
class TestClass; // forward declaration
namespace TestNamespace { // namespace body
void testFunc(TestClass &myObj); // passing by reference, ok!
}
class TestClass {
friend void TestNamespace::testFunc(TestClass &myObj); // scoping with namespace
};
Remember, namespace body can be only in global scope (not inside class or function).

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);
}