使用参数向量并行调用元素向量的成员函数

Question

鉴于这段代码：

struct T
{
  void f(int const);
};

void f(std::vector<T> &u, std::vector<int> const &v)
{
  for (std::size_t i = 0; i < u.size(); ++i)
    u[i].f(v[i]);
}

有没有标准的方法来并行化void f(std::vector<T> &u, std::vector<int> const &v) 的主体？

这碰巧奏效了 (https://godbolt.org/z/gRv9Ze)：

void f(std::vector<T> &u, std::vector<int> const &v)
{
  auto const indices = std::views::iota(0u, u.size()) | std::views::common;

  std::for_each(std::execution::par_unseq, std::begin(indices), std::end(indices),
                [&](std::size_t const i) { u[i].f(v[i]); });
}

但据报道，依赖这种行为是错误的（参见bug report 还有这个answer)。实际上，这不是并行运行的 (https://godbolt.org/z/MPGdHF)：

void f(std::vector<T> &u, std::vector<int> const &v)
{
  std::ranges::iota_view<std::size_t, std::size_t> const indices(0u, u.size());

  std::for_each(std::execution::par_unseq, std::begin(indices), std::end(indices),
                [&](std::size_t const i) { u[i].f(v[i]); });
}

我很确定应该有一种标准方法可以让类似的函数并行运行。我可能错过了一个明显的 算法，但是std::transform在这里似乎不合适，其他的更不合适。

Answer 1

留在std 内，最好的选择是std::transform，其输出迭代器会忽略给予它的内容

struct unit_iterator {
    using difference_type = std::ptrdiff_t;
    using value_type = std::tuple<>;
    using pointer = std::tuple<> *;
    using const_pointer = const std::tuple<> *;
    using reference = std::tuple<> &;
    using const reference = const std::tuple<> &;
    using iterator_category = std::random_access_iterator_tag;

    reference operator*() { return value; }
    const_reference operator*() const { return value; }
    reference operator[](difference_type) { return value; }
    const_reference operator[](difference_type) const { return value; }
    pointer operator->() { return &value; }
    const_pointer operator->() const { return &value; }

    unit_iterator& operator++() { return *this; }
    unit_iterator operator++(int) { return *this; }
    unit_iterator& operator+=(difference_type) { return *this; }
    unit_iterator operator+(difference_type) const { return *this; }

    unit_iterator& operator--() { return *this; }
    unit_iterator operator--(int) { return *this; }
    unit_iterator& operator-=(difference_type) { return *this; }
    unit_iterator operator-(difference_type) const { return *this; }

    difference_type operator-(unit_iterator) const { return 0; }

    bool operator==(unit_iterator) const { return true; }
    bool operator!=(unit_iterator) const { return false; }
    bool operator<(unit_iterator) const { return false; }
    bool operator<=(unit_iterator) const { return true; }
    bool operator>(unit_iterator) const { return false; }
    bool operator>=(unit_iterator) const { return true; }
private:
    static value_type value;
};


void f(std::vector<T> &u, std::vector<int> const &v)
{
  std::transform(std::execution::par_unseq, begin(u), end(u), begin(v), unit_iterator{},
                 [](T & u, int v) { u.f(v); return std::tuple<>{}; });
}