CODEWITHSUNDEEP
c++templates
user3684545
user3684545

Reputation: 45

template inherited class derived-base assignment

I have 2 classes, A and B and B is derived from A. I'm not allowed to change them. I'm implementing smart pointer class. 

template <typename T>
class SmartPtr{
    ...
    SmartPtr& operator=(const SmartPtr& p); 
    ...
}

template<typename T>
SmartPtr<T>& SmartPtr<T>::operator=(const SmartPtr & p) {
    ++*p.cnt;
    if(--*cnt == 0) { 
        delete ptr; 
        delete cnt; 
    }
    ptr = p.ptr;
    cnt = p.cnt;
    return *this;
}

I get error when I try to assign SmartPtr A to SmartPtr B as operand types are different. How can I fix that without changing A and B?

Upvotes: 0

Views: 125

Answers (1)

hellow
hellow

Reputation: 13430

your answer

I think you might mean SmartPtr<A> and SmartPtr<B>. Of course you won't be able to assign them, because class A is not class B. You will need to add another assignment operator, that looks like this

tempalte <typename U>
SmartPtr<T>& operator=(const SmartPtr<U>& p);

if you like to, you could use std::enable_if to make sure, U is derived from T

template <typename U>
SmartPtr<T>& operator=(const typename SmartPtr<std::enable_if<std::is_base_of<T, U>::value, U>::type &p);

and then you can implement this, by using dynamic_cast in your code to convert A to B

the 'correct' answer

using std::shared_ptr you could use this http://en.cppreference.com/w/cpp/memory/shared_ptr/pointer_cast

Upvotes: 1

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