Can't make operator<< in iterator class inside another class

Question

So, i have class ArrayList, and inside, i have class iterator. I tried everything to make operator<< inside iterator class, to work with iterators, but nothing works. I tried with and without friend, in private and public, inline and outside the class, i have nothing else to try so if someone has some suggestion, feel free to say :D tyyy

also, i don't know does it have something with my problem, but ArrayList is template class

template <typename T>
std::ostream& operator<<(std::ostream& os, const ArrayList<T>::iterator & it)
{
    return os << *it;
}

template <typename T>
class ArrayList {
private:
    T* elements_;
    size_t size_;
    size_t capacity_;
public:
class iterator;
};

template <typename T>
class ArrayList<T>::iterator {
private:
    T* ptr_;
public:
friend std::ostream& operator<<(std::ostream&, const iterator&);
};

This should be straightforward, so show us an example of what you have tried. You are obviously making some mistake, but it's hard to know what it is without seeing your code. — john, Apr 06 '20 at 06:47

NutCracker · Answer 1 · 2020-04-06T07:05:51.180

1

ArrayList<T>::iterator is a dependent name so it needs to be preceeded with typename keyword like:

template <typename T>
std::ostream& operator<<(std::ostream& os, const typename ArrayList<T>::iterator & it) {
//                                               ^^^
    return os << *it;
}

and, of course, your operator<< overload should be defined after your class definition.

Demo

UPDATE

In order to resolve linking errors mentioned in the comments, try placing the operator<< overload body directly into the class body like:

template <typename T>
class ArrayList<T>::iterator {
private:
    T* ptr_;
public:
    friend std::ostream& operator<<(std::ostream& os, const iterator& it) {
        return os << *it;
    }
};

edited Apr 06 '20 at 07:05

answered Apr 06 '20 at 06:53

NutCracker

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@vlasaks i believe that's the part you didn't show us? – NutCracker Apr 06 '20 at 07:00
@vlasaks at a guess probably due to https://stackoverflow.com/questions/495021/why-can-templates-only-be-implemented-in-the-header-file – Alan Birtles Apr 06 '20 at 07:00
@vlasaks try placing `operator< – NutCracker Apr 06 '20 at 07:02
ty kind sir, and everyone else the last one, inline worked, i tried the same thing without friend and it didn't, but now it's working :D ty again, i just spend 5h trying to solve it :D – vlasaks Apr 06 '20 at 07:15

armagedescu · Answer 2 · 2020-04-06T07:21:32.630

0

I would rather consider having operator *

template <typename T> class ArrayList {
private:
    T* elements_;
    size_t size_;
    size_t capacity_;
public:
    class iterator {
    private:
        T* ptr_;
    public:
        T operator *() {return *ptr_;}
    };
};

Now the operator << will work naturally for any type T that has operator defined, so you don't have to define it by yourself:

cout<< *it<< endl;

You will define the operator << only on typename T if it does not have operator defined. Normally you will not define it for int, char, string and so on.

edited Apr 06 '20 at 07:21

answered Apr 06 '20 at 07:16

armagedescu

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but i need operator << to work directly with iterator – vlasaks Apr 06 '20 at 07:21
How exactly you need it to work directly with iterator? What exactly the operator has to do with iterator? – armagedescu Apr 06 '20 at 07:23
@vlasaks the iterator implements the paradigm of pointer. A pointer need such operators as ++ -- * -> + .... But streaming a pointer with << is not a correct operation with a pointer. – armagedescu Apr 06 '20 at 07:28
@vlasaks You have to see what the iterators do. The std containers iterators do not implement such an operator. – armagedescu Apr 06 '20 at 07:30

Can't make operator<< in iterator class inside another class

2 Answers2