CODEWITHSUNDEEP
c++inheritanceoperator-overloading
user25470
user25470

Reputation: 595

trouble with operator overloading in baseclass

Hi I'm having some trouble with inhertance and operator overloading and I'm hoping you guys can give me some clarity.

I have the following classes:

template<typename Type>
class Predicate{
public:
    Predicate() {};
    virtual ~Predicate(){};
    virtual bool operator()(const Type & value) = 0;
    virtual bool operator()(const Type * value){ //<-- this is the operator thats not working
        return (*this)(*value);
    };
};

template<typename Type>
class Always : public Predicate<Type>{
public:
    bool operator()(const Type & value){return true;}
    ~Always(){};
};

Now I want all my predicates to accept both references and pointers, but when I test the classes in:

int main(){
    Always<int> a;
    int i = 1000;
    a(&i);
    system("pause");
    return 1;
}

I receive the following error:

test.cpp: In function 'int main()':
test.cpp:10:6: error: invalid conversion from 'int*' to 'int' [-fpermissive]
  a(&i);
      ^
In file included from test.cpp:2:0:
predicates.h:22:7: error:   initializing argument 1 of 'bool Always<Type>::operator()(const Type&) [with Type = int]' [-fpermissive]
  bool operator()(const Type & value){return true;}

Upvotes: 1

Views: 116

Answers (2)

Christian Hackl
Christian Hackl

Reputation: 27528

Templates and operator overloading obfuscate the real problem here. Look at this small piece of code which yields the same error:

void f(int &);

int main()
{
  int *ptr;
  f(ptr);
}

The compiler won't let you pass a pointer where a reference is expected. This is what you try to do with your derived class. As you operate on a concrete Always, the base versions of operator() are not considered.

Look how the situation changes when you operate instead on a pointer (or reference) to the base class:

int main(){
    Predicate<int> *ptr = new Always<int>;
    int i = 1000;
    (*ptr)(&i);
    delete ptr;
}

This compiles fine because the base-class operators are now considered for overload resolution. But this is just to make you understand the problem better. The solution is to apply the Non-Virtual Interface Idiom. Make your operators non-virtual and implement them in terms of private virtual functions:

template<typename Type>
class Predicate{
public:
    Predicate() {};
    virtual ~Predicate(){};
    bool operator()(const Type & value) { return operatorImpl(value); }
    bool operator()(const Type * value) { return operatorImpl(value); }

private:
    virtual bool operatorImpl(const Type & value) = 0;
    virtual bool operatorImpl(const Type * value) {
        return (*this)(*value);
    }
};

template<typename Type>
class Always : public Predicate<Type>{
public:
    ~Always(){};
private:
    bool operatorImpl(const Type & value){return true;}
};

Upvotes: 0

Shoe
Shoe

Reputation: 76240

This is because when you are declaring:

bool operator()(const Type & value){return true;}

in the subclass, you are hiding/shadowing any other overload of the operator in the superclass.

If you add:

using Predicate<Type>::operator();

Live demo

in the subclass, everything will work fine.

On a side note, I think that allowing both const& and const* is a design smell. You should just allow the const& version and let the user of your class do *ptr if they have a ptr pointer.

Upvotes: 2

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