检查值是否属于某个静态集的最C ++方法是什么？ [英] What's the most C++ way to check if value belongs to certain static set?

问题描述

假设我要编写这样的内容（ {1、3、7、42、69、550123} 集在编译前就已经知道）：

Let's say that I want to write something like this (the {1, 3, 7, 42, 69, 550123} set is known before compilation):

int x;
...
if (x == 1 || x == 3 || x == 7 || x == 42 || x == 69 || x == 5550123)
{
  ...
}

条件看起来很丑，因为我们每个可能的值都有9个额外的符号（ || x == ）。我该如何用更C ++的方式重写它？

The condition looks ugly because we have 9 extra symbols ("|| x ==") for each possible value. How can I rewrite it in a more C++ way?

我最好的猜测是：

int x;
...
const std::unordered_set<int> v = {1, 3, 7, 42, 69, 5550123};
if (v.count(x))
{
  ...
}

它的平均复杂度为O（1），有一些内存和时间开销，但看起来仍然很丑。

It has O(1) average complexity with some memory and time overhead, but still looks kinda ugly.

推荐答案

编辑：我刚刚注意到c ++ 14标记。请注意，我的 in 的实现依赖于C ++ 17。也可以使用递归在C ++ 14中完成，但这涉及更多样板，并且编译速度稍慢。

I just noticed c++14 tag. Note that my implementation of in relies on C++17. It can be done in C++14 as well using recursion, but that involves much more boilerplate, and a bit slower compilation.

一个人可以使用模板来生成一个函数具有逻辑运算符序列，例如nvoigt答案中的一个：

One could use a template to generate a function with a logical operator sequence, such as the one in nvoigt's answer:

template<auto... ts, class T>
constexpr bool
in(const T& t) noexcept(noexcept(((t == ts) || ...))) {
    return ((t == ts) || ...);
}

// usage
if (in<1, 3, 7, 42, 69, 5550123>(x))

也就是说，将一组魔术数字隐藏在命名函数的后面可能很有意义：

That said, hiding the set of magic numbers behind a named function probably makes a lot of sense:

constexpr bool
is_magical(int x) noexcept {
    return in<1, 3, 7, 42, 69, 5550123>(x);
}

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