iterators in c++ template class

Question

I have a little problem and I'm working on it for several hours but can't find a solution. Hope you will help me.

Here is my class:

#include <iostream>
#include <iterator>

template <typename T> class Array{
private:
    size_t size;
    T *newArray;

public:
    class Iterator:public std::iterator<std::output_iterator_tag, T>{
        T *p;
        public:
            Iterator(T*x):p(x){}
            T& operator*() {return *p;}

    };

    Array (size_t size = 10): size(size), newArray(new T[size]){};
    Iterator begin(){return (Iterator(newArray));}

    T printBegin(typename Array<T>::Iterator it){ return *it;}

    template <typename E>
    T printBegin(typename Array<E>::Iterator it){ return (T)*it;}

};

And here is Main:

using namespace std;

int main(){

    Array<int> x;
    Array<int> y;
    cout << y.printBegin(x.begin()); // prints 0 OK

    Array<double> p;
    // cout << p.printBegin(x.begin());

    return 0;
}

The first cout works fine but the line that is commented gives this : error: no matching function for call to ‘Array<double>::printBegin(Array<int>::Iterator)’

I don't understand because the last line in my Array class matches (normally) for this function call

I found this answer on another question that describes issues with trying to automatically deduce the a template function argument using a nested class (such as your Array<E>::Iterator). Also note that you can explicitly specify the template argument and it will work - cout << p.printBegin<int>(x.begin());. — Peter Clark
– Peter Clark, Commented Aug 11, 2014 at 13:44
It seems, that E cannot be deduced automatically. Try it this way: cout << p.printBegin<int>(x.begin()); — Simon
– Simon, Commented Aug 11, 2014 at 13:44
possible duplicate of Why is the template argument deduction not working here? — n. m. could be an AI
– n. m. could be an AI, Commented Aug 11, 2014 at 13:55
Your class doesn't make any sense whatsoever. Why should the Array class know how to print itself? Why should it be able to print other arrays other than its own? Why are you not using std::array or std::vector together with std::copy/std::back_inserter or a simple for loop to print the elements out on std::cout? Do you have a Java background? — sh03
– sh03, Commented Aug 11, 2014 at 14:01

rodrigo · Accepted Answer · 2014-08-11 14:13:28Z

1

The problem is that you want to deduce the E in Array<E>::Iterator from x.begin() that is an Array<int>::Iterator. But that is simply not possible.

Your best option is probably to write:

template <typename IT>
T printBegin(IT it){ return (T)*it;}

~~If, for any reason, you need to use the E type, then it is better to add a nested typedef:~~

class Iterator:public std::iterator<std::output_iterator_tag, T>{
    public:
        typename T array_member_type;
        //...
};

And then:

template <typename IT>
T printBegin(IT it){
    typedef typename IT::array_member_type E;
    return (T)*it;
}

Wait! Your iterator derives from std::iterator so the typedef already exists. No need to redefine:

template <typename IT>
T printBegin(IT it){
    typedef typename IT::value_type E;
    return (T)*it;
}

edited Aug 11, 2014 at 14:13

answered Aug 11, 2014 at 13:59

rodrigo

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Comments

CashCow · Accepted Answer · 2014-08-11 14:04:12Z

0

Your question is very similar to another question about template deduction.

The compiler cannot deduct the type E from Array<E>::Iterator. Why should it be able to? Two different types E could share the same iterator type.

What you are really saying to the compiler is:

"Is there a type "E" for which Array<E>::Iterator is this type?"

answered Aug 11, 2014 at 14:04

CashCow

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