Question

I have redefined the << operator and I want it to take a reference of pointer.

class foo
{
    foo();
    virtual ~foo();

    void operator << (BaseService*& iRight);
}

Somewhere in the code, having a foo instance, and a service which is a specialization of the BaseService class I do :

Service* service_pointer = new Service();
foo_instance << service_pointer;

But I get this error :
error: no match for ‘operator<<‘in ‘foo_instance << service_pointer’
note: candidates are: void foo::operator<<(BaseService*&)

Nothing changes if I dynamic_cast my service_pointer to BaseService

Service* service_pointer = new Service();
foo_instance << dynamic_cast<BaseService*>(service_pointer);

Any idea ?

Answer 1

The first version does not work because you can’t pass in a reference to a pointer to a subtype and rightly so: What if the implementation of operator<< made the pointer point to an instance MyService (which is a subclass of BaseService, but not of Service)? Clearly it would be illegal for a Service* to point to a MyService. So passing in a Service* is not allowed.

The second version is not allowed because dynamic_cast does not return an l-value, so you can’t pass it as a non-const reference.

The only thing you can do is define another variable of type BaseService* and pass that as an argument to <<. If << then reassigns the pointer, that change will be visible for the newly created variable only and not affect ServicePointer.

That being said (and not knowing your use case) I have to advice you that having operator<< take a non-const reference to anything as its right operand strikes me as bad practice. You wouldn’t usually expect << to modify it’s right operand.