Protected Base Destructor
Hello
I want to create a derived object that can decide weather or not it gets deleted.
this is what I tried, and it works, but it shows a warning:
warning: deleting object of polymorphic class type 'derived' which has non-virtual destructor might cause undefined behaviour.
So, is this safe?
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#include <iostream>
class base{
public:
base(){
std::cout << "creating base\n";
}
virtual bool kill()=0; //returns false if object should not be deleted for some reason
protected:
~base(){
std::cout << "destroying base\n";
}
};
class derived:public base{
public:
derived(): base(){
std::cout << "creating derived\n";
}
bool kill(){
delete this;
}
protected:
~derived(){
std::cout << "destroying derived\n";
}
};
int main(){
base* pBase= new derived;
pBase->kill();
return 0;
}
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thanks
Last edited on
No:
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#include <iostream>
class base{
public:
base(){
std::cout << "creating base\n";
}
virtual bool kill()=0; //returns false if object should not be deleted for some reason
protected:
~base(){
std::cout << "destroying base\n";
}
};
class derived:public base{
public:
derived(): base(){
std::cout << "creating derived\n";
}
bool kill(){
delete this;
}
protected:
~derived(){
std::cout << "destroying derived\n";
}
};
class derived2:public derived{
public:
derived2() {
std::cout << "creating derived2\n";
}
protected:
~derived2(){
std::cout << "destroying derived2\n";
}
};
int main(){
base* pBase= new derived2;
pBase->kill();
return 0;
}
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creating base
creating derived
creating derived2
destroying derived
destroying base |
'undefined behaviour' is always a bad thing
OK thanks didn't think of that.
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