error: deduced class type 'tuple' in function return type

What I am doing

I am practicing c++ after 3 years. I needed to learn fast and broadly, so this example i am trying to solve might look odd to you.

I am using c++20, gcc 10.2.

I wanted to make a pythonic enumerate function that

• Takes any container<T>
• Yields std::tuple<int, T>
• Where T is the type of items in the container

I wanted to try applying pythonic range as an argument of enumerate which

• Takes (int start, int end, int step)
• Yields int i from start to end every step

range (Not my code, I only added step functionality)

template <typename T>
class range_iterator;

template <typename T>
class range_impl
{
    const T start_;
    const T stop_;
    const T step_;

public:
    range_impl(T start, T stop, T step) : start_{start}, stop_{stop}, step_{step} {};
    range_impl(T start, T stop) : start_{start}, stop_{stop}, step_{1} {};
    range_impl(T stop) : start_{0}, stop_{stop}, step_{1} {};

    range_iterator<T> begin() const
    {
        return range_iterator<T>{start_, step_};
    }

    range_iterator<T> end() const
    {
        return range_iterator<T>{stop_, step_};
    }
};

template <typename T>
class range_iterator
{
    T current_;
    const T step_;

public:
    range_iterator(T init, T step) : current_{init}, step_{step} {};

    range_iterator<T> &operator++()
    {
        current_ += step_;
        return *this;
    }

    bool operator!=(const range_iterator<T> &rhs) const
    {
        return current_ != rhs.current_;
    }

    T operator*() const
    {
        return current_;
    }
};

template <typename T>
range_impl<T> range(const T start, const T stop, const T step)
{
    return range_impl<T>(start, stop, step);
}

template <typename T>
range_impl<T> range(const T start, const T stop)
{
    return range_impl<T>(start, stop);
}

template <typename T>
range_impl<T> range(const T stop)
{
    return range_impl<T>(stop);
}

Can be used like following

#include <iostream>
int main()
{
    for(auto i: range(0, 100 2)
    {
        std::cout << i << std::endl;
    }
}

Problem code: enumerate

template <typename T>
class enumerate_iterator;

// Here, T should be a type of a container that contains type X 
template <typename T>
class enumerate_impl
{
    T impl;

public:
    enumerate_impl<T>(T impl) : impl{impl} {/* empty */};

    enumerate_iterator begin() const
    {
        return enumerate_iterator{impl.begin()};
    }

    enumerate_iterator end() const
    {
        return enumerate_iterator{impl.end()};
    }
};


// Here, T should be a type of a iterator, I think. Confused myself.
template <typename T>
class enumerate_iterator
{

    T iterator;
    int i;

public:
    enumerate_iterator(T iterator) : iterator{iterator}, i{0} {/* empty body */};

    enumerate_iterator<T> &operator++()
    {
        i++;
        iterator++;
        return *this;
    }

    bool operator!=(const enumerate_iterator<T> &rhs) const
    {
        return iterator != rhs.iterator;
    }

    std::tuple operator*() const
    {
        return {i, *iterator};
    }
};

template <typename T>
enumerate_impl<T> enumerate(T impl)
{
    return enumerate_impl<T>{impl};
}

Expected usage

#include <iostream>
int main()
{
    for (auto &[i, j] : enumerate(range(0, 100, 2)))
    {
        std::cout << i << " " << j << std::endl;
    }
}

Gotten error (There actually tons of compilation error, but I want to try more on the others).

utility.cpp:186:20: error: deduced class type 'tuple' in function return type
  186 |         std::tuple operator*() const

I guessed this is complaining that you didn't tell me what type the tuple contains. But the thing is, I don't know what type T iterator contains. How would I tell return type is std::tuple<int, type T contains>?

Thanks for reading this long long question.