当迭代器(输入参数)通常不是constexpr时，constexpr算法真的有用吗?

问题描述

在c ++ 20中提出，某些算法是constexpr.

proposed in c++ 20, some algorithms are constexpr.

例如:

template< class InputIt, class UnaryPredicate >
bool all_of( InputIt first, InputIt last, UnaryPredicate p );
(since C++11)
(until C++20)


template< class InputIt, class UnaryPredicate >
constexpr bool all_of( InputIt first, InputIt last, UnaryPredicate p );
(since C++20)

虽然我们知道迭代器通常不是constexpr.我认为这仅在constexpr容器的情况下有用.有人可以澄清我是否缺少什么，我的理解是否正确?.

While we know iterator are not constexpr generally. I think this is useful only in case of constexpr container. Can someone clarify if I am missing something and Whether my understanding is correct ?.

推荐答案

可以.让我们尝试另一种算法，据我所知，它不是C ++ 20中的 constexpr .但是定义一个 constexpr 版本并不难(我只是从 cppreference 复制了示例实现，并在其上拍了 constexpr ):

Sure it is. Let's try another algorithm, which is not yet constexpr in C++20 to my knowledge, std::iota. But it's not too hard to define a constexpr version (I just copied the sample implementation from cppreference and slapped constexpr on it):

template<class ForwardIterator, class T>
constexpr void my_iota(ForwardIterator first, ForwardIterator last, T value)
{
    while(first != last) {
        *first++ = value;
        ++value;
    }
}

那么有用吗?是的.只要创建迭代器作为评估常量表达式的一部分，算法的评估就可以出现在常量表达式中.例如:

So is it useful? Yes it is. So long as the iterators are created as part of evaluating a constant expression, the evaluation of the algorithm can appear in a constant expression. For instance:

template<std::side_t N, typename T>
constexpr make_iota_array(T start = {}) {
  std::array<T, N> ret{};
  my_iota(ret.begin(), ret.end(), start);
  return ret;
}

上面创建了一个使用iota算法初始化的数组.如果将函数作为评估常量表达式的一部分而调用，则对象 ret 将作为评估的一部分创建，其迭代器也将作为评估的一部分.这些是有效的:

The above creates an array initialized with the iota algorithm. If the function is called as part of evaluating a constant expression, the object ret is created as part of the evaluation, and so are its iterators. These are valid:

constexpr auto a1 = make_iota_array<10, int>();
constexpr auto a2 = make_iota_array<10>(0u);

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