I'm trying to derive a class from a Base that has a friend function defined. I want to create a friend function for my derived class that makes use of the Base's friend function, but preserves the type of my Derived class:
#include<algorithm>
class Base
{
public:
Base(int i, int j, int k)
{
vect[0] = i;
vect[1] = j;
vect[2] = k;
}
~Base();
friend Base myMin (const Base& argA,
const Base& argB);
protected:
// Member data
int vect[3];
};
class Derived : public Base
{
public:
Derived(int i, int j, int k)
:
Base(i,j,k)
{
}
~Derived();
friend Derived doSomething(const Derived& argA,
const Derived& argB);
};
Base
myMin (const Base& argA,
const Base& argB)
{
int i = std::min(argA.vect[0], argB.vect[0]);
int j = std::min(argA.vect[1], argB.vect[1]);
int k = std::min(argA.vect[2], argB.vect[2]);
return Base(i,j,k);
}
Derived doSomething(const Derived& argA,
const Derived& argB)
{
// Does other stuff too...
return myMin(argA, argB);
}
int main(int argc, char *argv[])
{
Derived testA(2,4,6);
Derived testB(3,3,7);
doSomething(testA,testB);
return 0;
}
I understand why this doesn't work (doSomething is given a Base to return, but is told to return Derived), and that this probably isn't very good C++ form. My Base class has a ton of friend functions similar to this, is there a way to make use of the friend functions with my Derived class without having to modify them?
Thank you
Edited: The error, for reference, is:
base.H: In function ‘Derived doSomething(const Derived&, const Derived&)’:
base.H:50:26: error: could not convert ‘myMin(const Base&, const Base&)((* &(& argB)->Derived::<anonymous>))’ from ‘Base’ to ‘Derived’
return myMin(argA, argB);
The most obvious way would be to change doSomething() to
Derived doSomething(const Derived& argA,
const Derived& argB)
{
Base temp(myMin(argA, argB));
Derived retval(temp.vect[0], temp.vect[1], temp.vect[2]);
return retval;
}
You'll probably also have to declare doSomething() to be a friend of Base as well.
Without knowing more about what you are REALLY trying to do, hard to advise further. The fact you need to do this at all is an example of a "code smell" - a positive sign that your design is flawed and, as your code keeps growing, you will keep needing to do workarounds like this. That gradually makes it harder and harder to change the code - each time you change one part, something else has to be worked around.
Related
I ran into an error yesterday and, while it's easy to get around, I wanted to make sure that I'm understanding C++ right.
I have a base class with a protected member:
class Base
{
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
This compiles and works just fine. Now I extend Base but still want to use b:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
Note that in this case DoSomething is still taking a reference to a Base, not Derived. I would expect that I can still have access to that.b inside of Derived, but I get a cannot access protected member error (MSVC 8.0 - haven't tried gcc yet).
Obviously, adding a public getter on b solved the problem, but I was wondering why I couldn't have access directly to b. I though that when you use public inheritance the protected variables are still visible to the derived class.
A class can only access protected members of instances of this class or a derived class. It cannot access protected members of instances of a parent class or cousin class.
In your case, the Derived class can only access the b protected member of Derived instances, not that of Base instances.
Changing the constructor to take a Derived instance will solve the problem.
protected members can be accessed:
through this pointer
or to the same type protected members even if declared in base
or from friend classes, functions
To solve your case you can use one of last two options.
Accept Derived in Derived::DoSomething or declare Derived friend to Base:
class Derived;
class Base
{
friend class Derived;
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
You may also consider public getters in some cases.
As mentioned, it's just the way the language works.
Another solution is to exploit the inheritance and pass to the parent method:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
Base::DoSomething(that);
d=0;
}
};
You have access to the protected members of Derived, but not those of Base (even if the only reason it's a protected member of Derived is because it's inherited from Base)
You can try with static_cast< const Derived*>(pBase)->Base::protected_member ...
class Base
{
protected:
int b;
public:
...
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b += static_cast<const Derived*>(&that)->Base::b;
d=0;
}
void DoSomething(const Base* that)
{
b += static_cast<const Derived*>(that)->Base::b;
d=0;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething()
{
b+=this->b;
d=0;
}
};
//this will work
Following the hack for stl I wrote a small code which seems to solve the problem of accessing the protected members in derived class
#include <iostream>
class B
{
protected:
int a;
public:
void dosmth()
{
a = 4;
}
void print() {std::cout<<"a="<<a<<std::endl;}
};
class D: private B
{
public:
void dosmth(B &b)
{
b.*&D::a = 5;
}
};
int main(int argc, const char * argv[]) {
B b;
D d;
b.dosmth();
b.print();
d.dosmth(b);
b.print();
return 0;
}
Prints
a=4
a=5
Use this pointer to access protected members
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
this->b+=that.b;
d=0;
}
};
This is a recurrent problem once again. Someone know a easy way to do that? Imagine I have the following:
class Base
{
public:
...
Base property(const std::string& name)=0;
};
class Derived:public Base
{
public:
Derived();
Derived(const Derived&& val);
Base property(const std::string& name)
{
Derived z;
return z;
}
}
There is a way for the Derived::property return being (internally) a Derived copy instead of only Base part copy, and with the Derived move constructor invoked?
May be a stupid question, but really I dont find solution. Why copy constructors on return dont copy the specialized class?
Thanks you!
You can't do this.
Returning by value conceptually (ignoring RVO and move semantics) means making a copy of whatever you return by using the copy constructor of the type which the function is declared to return. If you return a Derived, a copy of type Base will be made and you'll lose the Derived part of the object. This is known as slicing.
If you want to return a Derived object as a Base, you'll need to use pointers.
The only aproximation I can find for who search something similar (related with X3liF, TartanLlama and other responses)
#define overridable(T) ovr<T>
#define return_overload_allowed(TYPE) friend struct ovr<TYPE>; virtual void* clone() const
#define return_overload_basic_allowed(TYPE) friend struct ovr<TYPE>; virtual void* clone() const{return new TYPE(*this);}
template<typename T> struct ovr
{
T* _obj;
ovr(const T& t)
: _obj(reinterpret_cast<T*>(t.clone()))
{;}
ovr(ovr<T>&& v)
: _obj(v._obj)
{
v._obj=nullptr;
}
operator T&()
{
return *_obj;
}
virtual ~ovr()
{
delete _obj;
}
};
class BASE
{
return_overload_basic_allowed(BASE);
public:
virtual overridable(BASE) method1();
virtual ~BASE();
};
class DERIVED: public BASE
{
return_overload_basic_allowed(DERIVED);
public:
virtual overridable(BASE) method1()
{
DERIVED a;
return a;
}
virtual ~DERIVED();
};
DERIVED a;
auto x = a.method1();
BASE& really_derived = x;
This compiles fine. But don't meet practical and smart requiriments... :(
I ran into an error yesterday and, while it's easy to get around, I wanted to make sure that I'm understanding C++ right.
I have a base class with a protected member:
class Base
{
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
This compiles and works just fine. Now I extend Base but still want to use b:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
Note that in this case DoSomething is still taking a reference to a Base, not Derived. I would expect that I can still have access to that.b inside of Derived, but I get a cannot access protected member error (MSVC 8.0 - haven't tried gcc yet).
Obviously, adding a public getter on b solved the problem, but I was wondering why I couldn't have access directly to b. I though that when you use public inheritance the protected variables are still visible to the derived class.
A class can only access protected members of instances of this class or a derived class. It cannot access protected members of instances of a parent class or cousin class.
In your case, the Derived class can only access the b protected member of Derived instances, not that of Base instances.
Changing the constructor to take a Derived instance will solve the problem.
protected members can be accessed:
through this pointer
or to the same type protected members even if declared in base
or from friend classes, functions
To solve your case you can use one of last two options.
Accept Derived in Derived::DoSomething or declare Derived friend to Base:
class Derived;
class Base
{
friend class Derived;
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
You may also consider public getters in some cases.
As mentioned, it's just the way the language works.
Another solution is to exploit the inheritance and pass to the parent method:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
Base::DoSomething(that);
d=0;
}
};
You have access to the protected members of Derived, but not those of Base (even if the only reason it's a protected member of Derived is because it's inherited from Base)
You can try with static_cast< const Derived*>(pBase)->Base::protected_member ...
class Base
{
protected:
int b;
public:
...
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b += static_cast<const Derived*>(&that)->Base::b;
d=0;
}
void DoSomething(const Base* that)
{
b += static_cast<const Derived*>(that)->Base::b;
d=0;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething()
{
b+=this->b;
d=0;
}
};
//this will work
Following the hack for stl I wrote a small code which seems to solve the problem of accessing the protected members in derived class
#include <iostream>
class B
{
protected:
int a;
public:
void dosmth()
{
a = 4;
}
void print() {std::cout<<"a="<<a<<std::endl;}
};
class D: private B
{
public:
void dosmth(B &b)
{
b.*&D::a = 5;
}
};
int main(int argc, const char * argv[]) {
B b;
D d;
b.dosmth();
b.print();
d.dosmth(b);
b.print();
return 0;
}
Prints
a=4
a=5
Use this pointer to access protected members
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
this->b+=that.b;
d=0;
}
};
I have searched far and wide for a specific answer to this question, and cannot find it. I am trying to create a base class with a virtual operator> that I can override in the derived class. Currently I'm having problems because declaring the function only requires one input variable (as in "bool operator> (Derived & a)" but attempting to define it in a cpp file tells me that it requires two inputs (as in "bool operator> (Derived & a, Derived & b))
I've tried defining the operator inline, but then I get errors where it thinks the derived class is still abstract because I'm passing in the derived type to the operator as shown above, instead of the base class. But if I pass the base class, then I cannot access the derived member variables I need to make the comparison.
I think I'm missing something simple here but I cannot seem to figure out what it is.
Hopefully you can help.
Thanks
For virtual calls to work from a reference/pointer of the base, you will need to use the base-type in the function, so for example
class Derived : public Base
{
....
bool operator>(Base &a)
{
Derived *pa = dynamic_cast<Derived *>(&a);
return this->something > pa->something; // Or whatever...
}
....
};
If you change the type, it becomes a different function, and when you use the base pointer or reference to refer to operator>, it will use the one in the base-class.
Why don't you leave operator>() non-virtual, und have it call a private virtual function?
Like so:
class Base {
public:
bool operator>(Base &a) {
return implementingFunction(a);
}
private:
virtual bool implementingFunction(Base &a) = 0;
};
#include <iostream>
using namespace std;
class base{
public :
virtual bool operator> (base& obj) { cout<<"b\n";return true;}
virtual ~base(){}
};
class derived: public base{
public:
virtual bool operator> (derived& obj) { cout<<"d\n";return true;}
~derived(){}
};
int main()
{
base *a=new derived(),b;
if(*a>b) { delete a; cout<<"Done!\n"; }
return 0;
}
Old question, but I've hardly seen a useful/correct answer here, so I would add my suggestion:
struct base
{
virtual ~base() = default;
virtual bool operator> (base const& obj) const = 0;
};
struct derived: public base
{
derived(int member) : member(member) {}
int member = 0;
virtual bool operator> (base const& obj) const
{
return member > static_cast<derived const&>(obj).member;
}
};
int main()
{
//in reality one would use a unique_ptr, of course
base* a = new derived(1);
base* b = new derived(0);
if(*a > *b)
{
//do something
}
return 0;
}
Caution: this works safely only if you're sure that the base const& parameter is really a derived const& (as e.g. in CRTP).
If not, you should use a dynamic_cast and add some error handling.
I ran into an error yesterday and, while it's easy to get around, I wanted to make sure that I'm understanding C++ right.
I have a base class with a protected member:
class Base
{
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
This compiles and works just fine. Now I extend Base but still want to use b:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
Note that in this case DoSomething is still taking a reference to a Base, not Derived. I would expect that I can still have access to that.b inside of Derived, but I get a cannot access protected member error (MSVC 8.0 - haven't tried gcc yet).
Obviously, adding a public getter on b solved the problem, but I was wondering why I couldn't have access directly to b. I though that when you use public inheritance the protected variables are still visible to the derived class.
A class can only access protected members of instances of this class or a derived class. It cannot access protected members of instances of a parent class or cousin class.
In your case, the Derived class can only access the b protected member of Derived instances, not that of Base instances.
Changing the constructor to take a Derived instance will solve the problem.
protected members can be accessed:
through this pointer
or to the same type protected members even if declared in base
or from friend classes, functions
To solve your case you can use one of last two options.
Accept Derived in Derived::DoSomething or declare Derived friend to Base:
class Derived;
class Base
{
friend class Derived;
protected:
int b;
public:
void DoSomething(const Base& that)
{
b+=that.b;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b+=that.b;
d=0;
}
};
You may also consider public getters in some cases.
As mentioned, it's just the way the language works.
Another solution is to exploit the inheritance and pass to the parent method:
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
Base::DoSomething(that);
d=0;
}
};
You have access to the protected members of Derived, but not those of Base (even if the only reason it's a protected member of Derived is because it's inherited from Base)
You can try with static_cast< const Derived*>(pBase)->Base::protected_member ...
class Base
{
protected:
int b;
public:
...
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
b += static_cast<const Derived*>(&that)->Base::b;
d=0;
}
void DoSomething(const Base* that)
{
b += static_cast<const Derived*>(that)->Base::b;
d=0;
}
};
class Derived : public Base
{
protected:
int d;
public:
void DoSomething()
{
b+=this->b;
d=0;
}
};
//this will work
Following the hack for stl I wrote a small code which seems to solve the problem of accessing the protected members in derived class
#include <iostream>
class B
{
protected:
int a;
public:
void dosmth()
{
a = 4;
}
void print() {std::cout<<"a="<<a<<std::endl;}
};
class D: private B
{
public:
void dosmth(B &b)
{
b.*&D::a = 5;
}
};
int main(int argc, const char * argv[]) {
B b;
D d;
b.dosmth();
b.print();
d.dosmth(b);
b.print();
return 0;
}
Prints
a=4
a=5
Use this pointer to access protected members
class Derived : public Base
{
protected:
int d;
public:
void DoSomething(const Base& that)
{
this->b+=that.b;
d=0;
}
};