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How to create a `range`-like iterable object of floats?

I want to create a range-like construct in c++, that will be used like this:

for (auto i: range(5,9))
    cout << i << ' ';    // prints 5 6 7 8 

for (auto i: range(5.1,9.2))
    cout << i << ' ';    // prints 5.1 6.1 7.1 8.1 9.1

Handling the integer case is relatively easy:

template<typename T>
struct range 
{
    T from, to;
    range(T from, T to) : from(from), to(to) {}

    struct iterator
    {
        T current;
        T operator*() {  return current; }

        iterator& operator++()
        {
            ++current;
            return *this;
        }

        bool operator==(const iterator& other) { return current == other.current; }
        bool operator!=(const iterator& other) { return current != other.current; }
    };

    iterator begin() const { return iterator{ from }; }
    iterator end()   const { return iterator{ to }; }
};

However, this does not work in the float case, since the standard range-based loop in C++ checks whether iter==end and not whether iter <= end as you would do in a for a loop.

Is there a simple way to create an iterable object that will behave like a correct range based for-loop on floats?

like image 374
Erel Segal-Halevi Avatar asked May 20 '19 09:05

Erel Segal-Halevi


3 Answers

Here is my attempt which does not impair the semantics of iterators. Now, each iterator knows its stopping value. The iterator will set itself to this value upon exceeding it. All end iterators of a range with equal to therefore compare equal.

template <typename T> 
struct range {
    T from, to;
    range(T from, T to): from(from), to(to) {}

    struct iterator {
        const T to; // iterator knows its bounds
        T current;

        T operator*() { return current; }

        iterator& operator++() { 
            ++current;
            if(current > to)
                // make it an end iterator
                // (current being exactly equal to 'current' of other end iterators)
                current = to;
            return *this;
        }

        bool operator==(const iterator& other) const // OT: note the const
        { return current == other.current; }
        // OT: this is how we do !=
        bool operator!=(const iterator& other) const { return !(*this == other); }
    };

    iterator begin() const { return iterator{to, from}; }
    iterator end()   const { return iterator{to, to}; }
};

Why is this better?

The solution by @JeJo relies on the order in which you compare those iterators, i.e. it != end or end != it. But, in the case of range-based for, it is defined. Should you use this contraption in some other context, I advise the above approach.


Alternatively, if sizeof(T) > sizeof(void*), it makes sense to store a pointer to the originating range instance (which in the case of the range-for persists until the end) and use that to refer to a single T value:

template <typename T> 
struct range {
    T from, to;
    range(T from, T to): from(from), to(to) {}

    struct iterator {
        range const* range;
        T current;

        iterator& operator++() { 
            ++current;
            if(current > range->to)
                current = range->to;
            return *this;
        }

        ...
    };

    iterator begin() const { return iterator{this, from}; }
    iterator end()   const { return iterator{this, to}; }
};

Or it could be T const* const pointing directly to that value, it is up to you.

OT: Do not forget to make the internals private for both classes.

like image 72
LogicStuff Avatar answered Nov 15 '22 13:11

LogicStuff


Instead of a range object you could use a generator (a coroutine using co_yield). Despite it is not in the standard (but planned for C++20), some compilers already implement it.

See: https://en.cppreference.com/w/cpp/language/coroutines

With MSVC it would be:

#include <iostream>
#include <experimental/generator>

std::experimental::generator<double> rangeGenerator(double from, double to) {
    for (double x=from;x <= to;x++)
    {
        co_yield x;
    }
}

int main()
{
    for (auto i : rangeGenerator(5.1, 9.2))
        std::cout << i << ' ';    // prints 5.1 6.1 7.1 8.1 9.1
}
like image 35
P. PICARD Avatar answered Nov 15 '22 12:11

P. PICARD


Is there a simple way to create an iterable object that will behave like a correct for loop on floats?

The simplest hack would be using the traits std::is_floating_point to provide different return (i.e. iter <= end) within the operator!= overload.

(See Live)

#include <type_traits>

bool operator!=(const iterator& other)
{
    if constexpr (std::is_floating_point_v<T>) return current <= other.current;
    return !(*this == other);
}

Warning: Even though that does the job, it breaks the meaning of operator!= overload.


Alternative Solution

The entire range class can be replaced by a simple function in which the values of the range will be populated with the help of std::iota in the standard container std::vector.

Use SFINE, to restrict the use of the function for only the valid types. This way, you can rely on standard implementations and forget about the reinventions.

(See Live)

#include <iostream>
#include <type_traits>
#include <vector>      // std::vector
#include <numeric>     // std::iota
#include <cstddef>     // std::size_t
#include <cmath>       // std::modf

// traits for valid template types(integers and floating points)
template<typename Type>
using is_integers_and_floats = std::conjunction<
    std::is_arithmetic<Type>,
    std::negation<std::is_same<Type, bool>>,
    std::negation<std::is_same<Type, char>>,
    std::negation<std::is_same<Type, char16_t>>,
    std::negation<std::is_same<Type, char32_t>>,
    std::negation<std::is_same<Type, wchar_t>>
    /*, std::negation<std::is_same<char8_t, Type>> */ // since C++20
>;    

template <typename T>
auto ragesof(const T begin, const T end)
               -> std::enable_if_t<is_integers_and_floats<T>::value, std::vector<T>>
{
    if (begin >= end) return std::vector<T>{}; // edge case to be considered
    // find the number of elements between the range
    const std::size_t size = [begin, end]() -> std::size_t 
    {
        const std::size_t diffWhole
                 = static_cast<std::size_t>(end) - static_cast<std::size_t>(begin);
        if constexpr (std::is_floating_point_v<T>) {
            double whole; // get the decimal parts of begin and end
            const double decimalBegin = std::modf(static_cast<double>(begin), &whole);
            const double decimalEnd   = std::modf(static_cast<double>(end), &whole);
            return decimalBegin <= decimalEnd ? diffWhole + 1 : diffWhole;
        }
        return diffWhole;
    }();
    // construct and initialize the `std::vector` with size
    std::vector<T> vec(size);
    // populates the range from [first, end)
    std::iota(std::begin(vec), std::end(vec), begin);
    return vec;
}

int main()
{
    for (auto i : ragesof( 5, 9 ))
        std::cout << i << ' ';    // prints 5 6 7 8
    std::cout << '\n';

    for (auto i : ragesof(5.1, 9.2))
            std::cout << i << ' '; // prints 5.1 6.1 7.1 8.1 9.1
}
like image 9
JeJo Avatar answered Nov 15 '22 11:11

JeJo