c++ compilation error - c++

i got a compile error which i do not understand.
i have a h/cpp file combination that does not contain a class but just defines some utility functions. when i try to use a struct that is defined in another class i get the error:
error C2027: use of undefined type 'B::C'
so, stripped down to the problem, the h-file looks like this
namespace A {
void foo(B::C::SStruct const & Var);
}
the definition of SStruct is in a class which is in another h-file, that is of course included.
namespace B {
class C {
public:
struct SStruct { };
};
}
the strange thing is, i can use this struct in other classes fine, it just seems to be related to this one h-file which contains just utility functions.
what am i missing here?
thanks!

After correcting missing semicolons etc. this compiles:
namespace B {
class C {
public:
struct SStruct { };
};
}
namespace A {
void foo(B::C::SStruct const & Var);
}
Obviously, if the order of the two namespaces were switched, this would not work. Possibly you are #including your headers in the wrong order. If this is the error, that's bad design - you should not allow header order to matter in your code.

I assume you meant "class C", not "Class C".
struct SStruct is private to class C (private being the default visibility of class members).
As such, it is not visible outside class C and its friends, which does not include A::foo().

Class C {
struct SStruct { };
}
==>
class C {
public:
struct SStruct { };
};

Your class is missing a semicolon after the definition. It should be:
namespace B {
class C {
public:
struct SStruct { };
}; // <=== Here
}

Related

Why use 'struct' keyword in class pointer declaration in C++

When and why should we use the 'struct' keyword when declaring a class pointer variable in C++?
I've seen this in embedded environments so I suspect that this is some kind of hold over from C. I've seen plenty of explanations on when to use the 'struct' keyword when declaring a struct object as it relates to namespaces in C (here), but I wasn't able to find anyone talking about why one might use it when declaring a class pointer variable.
Example, in CFoo.h:
class CFoo
{
public:
int doStuff();
};
inline Foo::doStuff()
{
return 7;
}
And later in a different class:
void CBar::interesting()
{
struct CFoo *pCFoo;
// Go on to do something interesting with pCFoo...
}
There's rarely a reason to do this: it's a fallover from C and in this case the programmer is simply being sentimental - perhaps it's there as a quest for readability. That said, it can be used in place of forward declarations.
In some instances you might need to disambiguate, but that's not the case here. One example where disambiguation would be necessary is
class foo{};
int main()
{
int foo;
class foo* pf1;
struct foo* pf2;
}
Note that you can use class and struct interchangeably. You can use typename too which can be important when working with templates. The following is valid C++:
class foo{};
int main()
{
class foo* pf1;
struct foo* pf2;
typename foo* pf3;
}
There are two reasons to do this.
The first one is if we are going to introduce a new type in the scope using an elaborated name. That is in this definition
void CBar::interesting()
{
struct CFoo *pCFoo;
// Go on to do something interesting with pCFoo...
}
the new type struct CFoo is introduced in the scope provided that it is not yet declared. The pointer may point to an incomplete type because pointers themselves are complete types.
The second one is when a name of a class is hidden by a declaration of a function or a variable. In this case we again need to use an elaborated type name.
Here are some examples
#include <iostream>
void CFoo( const class CFoo * c ) { std::cout << ( const void * )c << '\n'; }
class CFoo
{
public:
int doStuff();
};
int main()
{
class CFoo c1;
return 0;
}
Or
#include <iostream>
class CFoo
{
public:
int doStuff();
};
void CFoo( void ) { std::cout << "I am hidding the class CGoo!\n"; }
int main()
{
class CFoo c1;
return 0;
}
In C, two different styles are the most common:
typedef struct { ... } s; with variables declared as s name;.
struct s { ... }; with variables declared as struct s name;
In C++ you don't need to typedef to omit the struct keyword, so the former style is far more in line with the C++ type system and classes, making it the most common style in C++.
But then there are not many cases in C++ when you actually want to use struct instead of class in the first place - structs are essentially classes with all members public by default.
The reason for this may be as simple as not having to include a header file whose contents aren't needed other than for announcing that CFoo names a type. That's often done with a forward declaration:
class CFoo;
void f(CFoo*);
but it can also be done on the fly:
void f(struct CFoo*);

C++ Calling a function from another class

Very new to c++ having trouble calling a function from another class.
Class B inherits from Class A, and I want class A to be able to call a function created in class B.
using namespace std;
class B;
class A
{
public:
void CallFunction ()
{
B b;
b.bFunction();
}
};
class B: public A
{
public:
virtual void bFunction()
{
//stuff done here
}
};
It all looks fine on screen (no obvious errors) but when I try to compile it i get an error C2079 'b' uses undefined class B.
I've tried making them pointers/ friends but I'm getting the same error.
void CallFunction ()
{ // <----- At this point the compiler knows
// nothing about the members of B.
B b;
b.bFunction();
}
This happens for the same reason that functions in C cannot call each other without at least one of them being declared as a function prototype.
To fix this issue we need to make sure both classes are declared before they are used. We separate the declaration from the definition. This MSDN article explains in more detail about the declarations and definitions.
class A
{
public:
void CallFunction ();
};
class B: public A
{
public:
virtual void bFunction()
{ ... }
};
void A::CallFunction ()
{
B b;
b.bFunction();
}
What you should do, is put CallFunction into *.cpp file, where you include B.h.
After edit, files will look like:
B.h:
#pragma once //or other specific to compiler...
using namespace std;
class A
{
public:
void CallFunction ();
};
class B: public A
{
public:
virtual void bFunction()
{
//stuff done here
}
};
B.cpp
#include "B.h"
void A::CallFunction(){
//use B object here...
}
Referencing to your explanation, that you have tried to change B b; into pointer- it would be okay, if you wouldn't use it in that same place. You can use pointer of undefined class(but declared), because ALL pointers have fixed byte size(4), so compiler doesn't have problems with that. But it knows nothing about the object they are pointing to(simply: knows the size/boundary, not the content).
So as long as you are using the knowledge, that all pointers are same size, you can use them anywhere. But if you want to use the object, they are pointing to, the class of this object must be already defined and known by compiler.
And last clarification: objects may differ in size, unlike pointers. Pointer is a number/index, which indicates the place in RAM, where something is stored(for example index: 0xf6a7b1).
class B is only declared but not defined at the beginning, which is what the compiler complains about. The root cause is that in class A's Call Function, you are referencing instance b of type B, which is incomplete and undefined. You can modify source like this without introducing new file(just for sake of simplicity, not recommended in practice):
using namespace std;
class A
{
public:
void CallFunction ();
};
class B: public A
{
public:
virtual void bFunction()
{
//stuff done here
}
};
// postpone definition of CallFunction here
void A::CallFunction ()
{
B b;
b.bFunction();
}
in A you have used a definition of B which is not given until then , that's why the compiler is giving error .
Forward declare class B and swap order of A and B definitions: 1st B and 2nd A. You can not call methods of forward declared B class.
Here's my solution to the issue. Tried to keep it straight and simple.
#include <iostream>
using namespace std;
class Game{
public:
void init(){
cout << "Hi" << endl;
}
}g;
class b : Game{ //class b uses/imports class Game
public:
void h(){
init(); //Use function from class Game
}
}A;
int main()
{
A.h();
return 0;
}
You can also have a look at the curiously recurring template pattern and solve your problem similar to this:
template<typename B_TYPE>
struct A
{
int callFctn()
{
B_TYPE b;
return b.bFctn();
}
};
struct B : A<B>
{
int bFctn()
{
return 5;
}
};
int main()
{
A<B> a;
return a.callFctn();
}

template private inheritance in vc++10 is not accessible

The following code compiles using GCC 4.4.6 and Comeau 4.3.10.
#include <iostream>
struct A { int name; };
template<typename T> struct C : T { using T::name; };
struct B : private A { friend struct C<B>; };
int main()
{
C<B> o;
o.name = 0;
}
It gives the following error in VC++10:
main.cpp(4): error C2877: 'A::name' is not accessible from 'A'
main.cpp(10): error C2247: 'A::name' not accessible because 'B' uses 'private' to inherit from 'A'
What's a good cross-compiler workaround that allows o.name = 0;?
Note: Adding using A::name to B takes care of the problem, but publishes the A::name member to everyone, whereas it should only be visible to a particular template instantiation, namely C<B>.
Work around is what #kerrekSB suggested, add using A::name; in class B:
struct A { int name; };
template<typename T> struct C : T { using T::name; };
struct B : private A {
using A::name;
friend struct C<B>;
};
your initial example didn't work cause class A is private to B and class C<B> is friend of B but when you access member name from object of C<B> , line using T::name; creates problem since the class B doesn't has any member name in it. it's scope search which find the member name when you try to access it via object of class B
Edit :
Adding using A::name to B takes care of the problem, but publishes the
A::name member to everyone, whereas it should only be visible to a
particular template instantiation, namely C
if that's the case , then simply declare statement using A::name; in private section in class B i.e
struct B : private A {
protected: using A::name;
public:
friend struct C<B>;
};
There seems to be a fundamental difference in visibility considerations between gcc and VC++ when using member using-declarations; check this simplified example without templates:
struct A { int name; };
struct B: private A { friend struct C; };
struct C: B {using B::name; };
int main()
{
C o;
o.name = 0;
}
It will compile on gcc but not on VC++ (with basically the same error as in the question). Will have to consult the standard on who is doing it right...

Forward declare a inner class type, is it possible?

Ok here is the situation.
//foo.h
struct A1{
struct A2{};
};
//bar.h
#include "MyString.h"
class A2; //note, not including foo.h
TEMPLATE_INSTIANTIATE_MAP_OF_TYPE(String,A2*); //assume compiler doesn't do this
Is it possible to make the above situation work? I try to create a MyMap<String,A1::A2*> m; but the compilers throws undefined reference errors. Is it possible to make the above work without having bar.h import foo.h?
Sadly, it isn't. Nested classes can only be declared inside a class definition.
Here is a way to declare nested classes outside a class definition.
class Logic is the outer class. LogicImp is the forward declared struct.
class Logic
{
public:
Logic();
~Logic();
private:
struct LogicImp;
std::unique_ptr<LogicImp> limp_;
};
struct Logic::LogicImp
{
int nLogical_;
};
Logic::Logic():limp_(new LogicImp())
{
}
Logic::~Logic()
{
}

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);
}