I want to use one pointer ( _ref ) to point to different class types. In order to use it, I must cast it to the type that is addressed. I cannot do that because of the incomplete type that is at line 5. If i move the definition of B to line 5, it requires class A to be defined.
#include <iostream>
#include <vector>
#include <string>
class B;
class A{
void *_ref;
std::string _reft;
public:
void setref(A &a){
_ref=&a;
_reft=typeid(a).name();
}
void setref(B &b){
_ref=&b;
_reft=typeid(b).name();
}
void test(){
if(_ref && _reft==std::string(typeid(B).name())){
std::cout<<"Ref to B: ";
static_cast<B*>(_ref)->test(); //error here
}
}
};
class B{
std::vector<A> a;
public:
A A(int i){
return a[i];
}
void test(){
std::cout<<"IT WORKS!";
}
};
int main(){
A a;
B b;
a.setref(b);
a.test();
return 0;
}
Move the implementation of the function that requires B to be complete out of the class; put it either in a source file, or inline after the definition of B:
class A{
// ...
void test();
};
class B{
// ...
};
inline void A::test(){
// ...
}
If you use pointers rather than references you will be able to do this.
You need to change your function definitions to use pointers not references.
Then when calling the functions use the address of the object.
Related
I recently got stuck in a situation like this:
class A
{
public:
typedef struct/class {…} B;
…
C::D *someField;
}
class C
{
public:
typedef struct/class {…} D;
…
A::B *someField;
}
Usually you can declare a class name:
class A;
But you can't forward declare a nested type, the following causes compilation error.
class C::D;
Any ideas?
You can't do it, it's a hole in the C++ language. You'll have to un-nest at least one of the nested classes.
class IDontControl
{
class Nested
{
Nested(int i);
};
};
I needed a forward reference like:
class IDontControl::Nested; // But this doesn't work.
My workaround was:
class IDontControl_Nested; // Forward reference to distinct name.
Later when I could use the full definition:
#include <idontcontrol.h>
// I defined the forward ref like this:
class IDontControl_Nested : public IDontControl::Nested
{
// Needed to make a forwarding constructor here
IDontControl_Nested(int i) : Nested(i) { }
};
This technique would probably be more trouble than it's worth if there were complicated constructors or other special member functions that weren't inherited smoothly. I could imagine certain template magic reacting badly.
But in my very simple case, it seems to work.
If you really want to avoid #including the nasty header file in your header file, you could do this:
hpp file:
class MyClass
{
public:
template<typename ThrowAway>
void doesStuff();
};
cpp file
#include "MyClass.hpp"
#include "Annoying-3rd-party.hpp"
template<> void MyClass::doesStuff<This::Is::An::Embedded::Type>()
{
// ...
}
But then:
you will have to specify the embedded type at call time (especially if your function does not take any parameters of the embedded type)
your function can not be virtual (because it is a template)
So, yeah, tradeoffs...
I would not call this an answer, but nonetheless an interesting find:
If you repeat the declaration of your struct in a namespace called C, everything is fine (in gcc at least).
When the class definition of C is found, it seems to silently overwrite the namspace C.
namespace C {
typedef struct {} D;
}
class A
{
public:
typedef struct/class {...} B;
...
C::D *someField;
}
class C
{
public:
typedef struct/class {...} D;
...
A::B *someField;
}
If you have access to change the source code of classes C and D, then you can take out class D separately, and enter a synonym for it in class C:
class CD {
};
class C {
public:
using D = CD;
};
class CD;
This would be a workaround (at least for the problem described in the question -- not for the actual problem, i.e., when not having control over the definition of C):
class C_base {
public:
class D { }; // definition of C::D
// can also just be forward declared, if it needs members of A or A::B
};
class A {
public:
class B { };
C_base::D *someField; // need to call it C_base::D here
};
class C : public C_base { // inherits C_base::D
public:
// Danger: Do not redeclare class D here!!
// Depending on your compiler flags, you may not even get a warning
// class D { };
A::B *someField;
};
int main() {
A a;
C::D * test = a.someField; // here it can be called C::D
}
This can be done by forward declare the outer class as a namespace.
Sample: We have to use a nested class others::A::Nested in others_a.h, which is out of our control.
others_a.h
namespace others {
struct A {
struct Nested {
Nested(int i) :i(i) {}
int i{};
void print() const { std::cout << i << std::endl; }
};
};
}
my_class.h
#ifndef MY_CLASS_CPP
// A is actually a class
namespace others { namespace A { class Nested; } }
#endif
class MyClass {
public:
MyClass(int i);
~MyClass();
void print() const;
private:
std::unique_ptr<others::A::Nested> _aNested;
};
my_class.cpp
#include "others_a.h"
#define MY_CLASS_CPP // Must before include my_class.h
#include "my_class.h"
MyClass::MyClass(int i) :
_aNested(std::make_unique<others::A::Nested>(i)) {}
MyClass::~MyClass() {}
void MyClass::print() const {
_aNested->print();
}
Assuming I have these classes (question marks mark the question what I need to pass here):
class A
{
...
public:
void pass()
{
B ins;
ins.doSth(?????);
}
};
class B
{
...
public:
void doSth(const A &sth)
{
...
}
}
int main()
{
A te;
te.pass();
}
Can you pass an instance of your own class or is this just an example of a failed class structure on my side?
The current object in a member function is *this. You can pass that to another function.
You will have to consider how the classes depend on each other, and that one class cannot use the other class until the declaration is complete.
This would work though:
class A
{
//...
public:
void pass();
};
class B
{
//...
public:
void doSth(const A &sth)
{
//...
}
};
// Here both classes are completely declared
void A::pass()
{
B ins;
ins.doSth(*this);
}
int main()
{
A te;
te.pass();
}
Your class "contains" an instance of each other so you'll face an error of undeclared types. To solve this issue you need to use forward declaration.
And you'll face another problem:
If your methods doSth() and pass() are defined inlinlely then you'll face a problem: "using incomplete types". The workaround this is to implement these methods outside the class so that each object has been fully constructed before used.
The program may look like:
class A;
class B;
class A{
public:
void pass();
};
class B{
public:
void doSth(const A &sth){
}
};
void A::pass(){
B ins;
ins.doSth(*this);
}
int main(){
A te;
te.pass();
return 0;
}
I recently got stuck in a situation like this:
class A
{
public:
typedef struct/class {…} B;
…
C::D *someField;
}
class C
{
public:
typedef struct/class {…} D;
…
A::B *someField;
}
Usually you can declare a class name:
class A;
But you can't forward declare a nested type, the following causes compilation error.
class C::D;
Any ideas?
You can't do it, it's a hole in the C++ language. You'll have to un-nest at least one of the nested classes.
class IDontControl
{
class Nested
{
Nested(int i);
};
};
I needed a forward reference like:
class IDontControl::Nested; // But this doesn't work.
My workaround was:
class IDontControl_Nested; // Forward reference to distinct name.
Later when I could use the full definition:
#include <idontcontrol.h>
// I defined the forward ref like this:
class IDontControl_Nested : public IDontControl::Nested
{
// Needed to make a forwarding constructor here
IDontControl_Nested(int i) : Nested(i) { }
};
This technique would probably be more trouble than it's worth if there were complicated constructors or other special member functions that weren't inherited smoothly. I could imagine certain template magic reacting badly.
But in my very simple case, it seems to work.
If you really want to avoid #including the nasty header file in your header file, you could do this:
hpp file:
class MyClass
{
public:
template<typename ThrowAway>
void doesStuff();
};
cpp file
#include "MyClass.hpp"
#include "Annoying-3rd-party.hpp"
template<> void MyClass::doesStuff<This::Is::An::Embedded::Type>()
{
// ...
}
But then:
you will have to specify the embedded type at call time (especially if your function does not take any parameters of the embedded type)
your function can not be virtual (because it is a template)
So, yeah, tradeoffs...
I would not call this an answer, but nonetheless an interesting find:
If you repeat the declaration of your struct in a namespace called C, everything is fine (in gcc at least).
When the class definition of C is found, it seems to silently overwrite the namspace C.
namespace C {
typedef struct {} D;
}
class A
{
public:
typedef struct/class {...} B;
...
C::D *someField;
}
class C
{
public:
typedef struct/class {...} D;
...
A::B *someField;
}
If you have access to change the source code of classes C and D, then you can take out class D separately, and enter a synonym for it in class C:
class CD {
};
class C {
public:
using D = CD;
};
class CD;
This would be a workaround (at least for the problem described in the question -- not for the actual problem, i.e., when not having control over the definition of C):
class C_base {
public:
class D { }; // definition of C::D
// can also just be forward declared, if it needs members of A or A::B
};
class A {
public:
class B { };
C_base::D *someField; // need to call it C_base::D here
};
class C : public C_base { // inherits C_base::D
public:
// Danger: Do not redeclare class D here!!
// Depending on your compiler flags, you may not even get a warning
// class D { };
A::B *someField;
};
int main() {
A a;
C::D * test = a.someField; // here it can be called C::D
}
This can be done by forward declare the outer class as a namespace.
Sample: We have to use a nested class others::A::Nested in others_a.h, which is out of our control.
others_a.h
namespace others {
struct A {
struct Nested {
Nested(int i) :i(i) {}
int i{};
void print() const { std::cout << i << std::endl; }
};
};
}
my_class.h
#ifndef MY_CLASS_CPP
// A is actually a class
namespace others { namespace A { class Nested; } }
#endif
class MyClass {
public:
MyClass(int i);
~MyClass();
void print() const;
private:
std::unique_ptr<others::A::Nested> _aNested;
};
my_class.cpp
#include "others_a.h"
#define MY_CLASS_CPP // Must before include my_class.h
#include "my_class.h"
MyClass::MyClass(int i) :
_aNested(std::make_unique<others::A::Nested>(i)) {}
MyClass::~MyClass() {}
void MyClass::print() const {
_aNested->print();
}
I recently got stuck in a situation like this:
class A
{
public:
typedef struct/class {…} B;
…
C::D *someField;
}
class C
{
public:
typedef struct/class {…} D;
…
A::B *someField;
}
Usually you can declare a class name:
class A;
But you can't forward declare a nested type, the following causes compilation error.
class C::D;
Any ideas?
You can't do it, it's a hole in the C++ language. You'll have to un-nest at least one of the nested classes.
class IDontControl
{
class Nested
{
Nested(int i);
};
};
I needed a forward reference like:
class IDontControl::Nested; // But this doesn't work.
My workaround was:
class IDontControl_Nested; // Forward reference to distinct name.
Later when I could use the full definition:
#include <idontcontrol.h>
// I defined the forward ref like this:
class IDontControl_Nested : public IDontControl::Nested
{
// Needed to make a forwarding constructor here
IDontControl_Nested(int i) : Nested(i) { }
};
This technique would probably be more trouble than it's worth if there were complicated constructors or other special member functions that weren't inherited smoothly. I could imagine certain template magic reacting badly.
But in my very simple case, it seems to work.
If you really want to avoid #including the nasty header file in your header file, you could do this:
hpp file:
class MyClass
{
public:
template<typename ThrowAway>
void doesStuff();
};
cpp file
#include "MyClass.hpp"
#include "Annoying-3rd-party.hpp"
template<> void MyClass::doesStuff<This::Is::An::Embedded::Type>()
{
// ...
}
But then:
you will have to specify the embedded type at call time (especially if your function does not take any parameters of the embedded type)
your function can not be virtual (because it is a template)
So, yeah, tradeoffs...
I would not call this an answer, but nonetheless an interesting find:
If you repeat the declaration of your struct in a namespace called C, everything is fine (in gcc at least).
When the class definition of C is found, it seems to silently overwrite the namspace C.
namespace C {
typedef struct {} D;
}
class A
{
public:
typedef struct/class {...} B;
...
C::D *someField;
}
class C
{
public:
typedef struct/class {...} D;
...
A::B *someField;
}
If you have access to change the source code of classes C and D, then you can take out class D separately, and enter a synonym for it in class C:
class CD {
};
class C {
public:
using D = CD;
};
class CD;
This would be a workaround (at least for the problem described in the question -- not for the actual problem, i.e., when not having control over the definition of C):
class C_base {
public:
class D { }; // definition of C::D
// can also just be forward declared, if it needs members of A or A::B
};
class A {
public:
class B { };
C_base::D *someField; // need to call it C_base::D here
};
class C : public C_base { // inherits C_base::D
public:
// Danger: Do not redeclare class D here!!
// Depending on your compiler flags, you may not even get a warning
// class D { };
A::B *someField;
};
int main() {
A a;
C::D * test = a.someField; // here it can be called C::D
}
This can be done by forward declare the outer class as a namespace.
Sample: We have to use a nested class others::A::Nested in others_a.h, which is out of our control.
others_a.h
namespace others {
struct A {
struct Nested {
Nested(int i) :i(i) {}
int i{};
void print() const { std::cout << i << std::endl; }
};
};
}
my_class.h
#ifndef MY_CLASS_CPP
// A is actually a class
namespace others { namespace A { class Nested; } }
#endif
class MyClass {
public:
MyClass(int i);
~MyClass();
void print() const;
private:
std::unique_ptr<others::A::Nested> _aNested;
};
my_class.cpp
#include "others_a.h"
#define MY_CLASS_CPP // Must before include my_class.h
#include "my_class.h"
MyClass::MyClass(int i) :
_aNested(std::make_unique<others::A::Nested>(i)) {}
MyClass::~MyClass() {}
void MyClass::print() const {
_aNested->print();
}
I have a problem creating some form of hierarchy with different object types. I have a class which has a member of another class, like this:
class A
{
public:
A(){}
~A(){}
void addB(B* dep){
child = dep;
dep->addOwner(this);
}
void updateChild(){
child->printOwner();
}
void print(){
printf("Printing...");
}
private:
B* child;
};
And this is class B:
class B
{
public:
void addOwner(A* owner){
ownerObject = owner;
}
//ISNT WORKING
void printOwner(){
ownerObject->print();
}
private:
A* ownerObject;
};
Calling a function of "B" out of class "A" works just fine but trying it vice versa gives a compiler error because A is not defined in B. It actually is by using an include and a forward declaration, but I guess its a cross reference problem which the compiler can not solve.
Is there any chance to solve this problem or should I rethink my design?
You say that you already solved your circular dependency problem by using a forward declaration of A instead of including the header where A is defined, so you already know how to avoid circular includes. However, you should be aware of what is possible and what is not with incomplete types (i.e. types that have been forward declared).
In your case, you try to call the member function print on an object that has an incomplete type; the compiler knows nothing about this type excepts that it will be defined at some point, so it does not allow you to do this. The solution is to remove the implementation of the printOwner member function from the B header and put it into an implementation file:
//B.hpp
class A; // forward declaration
class B
{
public:
void addOwner(A* owner);
void printOwner() const; // I think this member function could be const
private:
A* ownerObject;
};
//B.cpp
#include "B.hpp"
#include "A.hpp" // here we "import" the definition of A
void B::addOwner(A * owner)
{
ownerObject = owner;
}
void B::printOwner() const
{
ownerObject->print(); //A is complete now, so we can use its member functions
}
You could possibly do the same thing in the A header.
You can use forward declaration, and define the member functions outside of the class, i.e.
// A.h
class B;
class A { public:
void addB(B* dep); // don't define addB here.
...
};
// B.h
class A;
class B { public:
void addOwner(A* owner); // don't define addOwner here.
...
};
// A.cpp
#include "A.h"
#include "B.h"
void A::addB(B* dep) {
...
}
// B.cpp
// similar.
You probably should rethink your design, since a crcular parent-child relationship is usually a code smell.
But, you can make the compiler happy :
#include <cstdlib>
#include <cstdio>
class A
{
public:
A(){}
~A(){}
void addB(class B* dep);
void updateChild();
void print(){
printf("Printing...");
}
private:
class B* child;
};
class B
{
public:
void addOwner(A* owner){
ownerObject = owner;
}
//ISNT WORKING
void printOwner(){
ownerObject->print();
}
private:
A* ownerObject;
};
void A::addB(class B* dep){
child = dep;
dep->addOwner(this);
}
void A::updateChild(){
child->printOwner();
}
int main()
{
return 0;
}
You should move B::printOwner implementation to .cpp file.