Difference between revisions of "cpp/algorithm/ranges/fold left first with iter"

Latest revision as of 07:57, 23 April 2024

Left-folds the elements of given range, that is, returns the result of evaluation of the chain expression:
f(f(f(f(x1, x2), x3), ...), xn), where x1, x2, ..., xn are elements of the range.

Informally, ranges::fold_left_first_with_iter behaves like std::accumulate's overload that accepts a binary predicate, except that the *first is used internally as an initial element.

The behavior is undefined if [first, last) is not a valid range.

The function-like entities described on this page are niebloids, that is:

• Explicit template argument lists cannot be specified when calling any of them.
• None of them are visible to argument-dependent lookup.
• When any of them are found by normal unqualified lookup as the name to the left of the function-call operator, argument-dependent lookup is inhibited.

In practice, they may be implemented as function objects, or with special compiler extensions.

[edit] Parameters

[edit] Return value

Let U be decltype(ranges::fold_left(std::move(first), last, std::iter_value_t<I>(*first), f)).

[edit] Possible implementations

class fold_left_first_with_iter_fn
{
    template<class O, class I, class S, class F>
    constexpr auto impl(I&& first, S&& last, F f) const
    {
        using U = decltype(
            ranges::fold_left(std::move(first), last, std::iter_value_t<I>(*first), f)
        );
        using Ret = ranges::fold_left_first_with_iter_result<O, std::optional<U>>;
        if (first == last)
            return Ret{std::move(first), std::optional<U>()};
        std::optional<U> init(std::in_place, *first);
        for (++first; first != last; ++first)
            *init = std::invoke(f, std::move(*init), *first);
        return Ret{std::move(first), std::move(init)};
    }
 
public:
    template<std::input_iterator I, std::sentinel_for<I> S,
             /*indirectly-binary-left-foldable*/<std::iter_value_t<I>, I> F>
    requires std::constructible_from<std::iter_value_t<I>, std::iter_reference_t<I>>
    constexpr auto operator()(I first, S last, F f) const
    {
        return impl<I>(std::move(first), std::move(last), std::ref(f));
    }
 
    template<ranges::input_range R, /*indirectly-binary-left-foldable*/<
        ranges::range_value_t<R>, ranges::iterator_t<R>> F>
    requires
        std::constructible_from<ranges::range_value_t<R>, ranges::range_reference_t<R>>
    constexpr auto operator()(R&& r, F f) const
    {
        return impl<ranges::borrowed_iterator_t<R>>(
            ranges::begin(r), ranges::end(r), std::ref(f)
        );
    }
};
 
inline constexpr fold_left_first_with_iter_fn fold_left_first_with_iter;

[edit] Complexity

Exactly ranges::distance(first, last) - 1 (assuming the range is not empty) applications of the function object f.

[edit] Notes

The following table compares all constrained folding algorithms:

[edit] Example

#include <algorithm>
#include <cassert>
#include <functional>
#include <iostream>
#include <ranges>
#include <utility>
#include <vector>
 
int main()
{
    std::vector v{1, 2, 3, 4, 5, 6, 7, 8};
 
    auto sum = std::ranges::fold_left_first_with_iter
    (
        v.begin(), v.end(), std::plus<int>()
    );
    std::cout << "sum: " << sum.value.value() << '\n';
    assert(sum.in == v.end());
 
    auto mul = std::ranges::fold_left_first_with_iter(v, std::multiplies<int>());
    std::cout << "mul: " << mul.value.value() << '\n';
    assert(mul.in == v.end());
 
    // get the product of the std::pair::second of all pairs in the vector:
    std::vector<std::pair<char, float>> data {{'A', 2.f}, {'B', 3.f}, {'C', 7.f}};
    auto sec = std::ranges::fold_left_first_with_iter
    (
        data | std::ranges::views::values, std::multiplies<>()
    );
    std::cout << "sec: " << sec.value.value() << '\n';
 
    // use a program defined function object (lambda-expression):
    auto lambda = [](int x, int y) { return x + y + 2; };
    auto val = std::ranges::fold_left_first_with_iter(v, lambda);
    std::cout << "val: " << val.value.value() << '\n';
    assert(val.in == v.end());
}

sum: 36
mul: 40320
sec: 42
val: 50

[edit] References

[edit] See also