C++ 中有标准的循环整数类吗？

发布于 2024-09-24 21:30:09 字数 616 浏览 8 评论 0原文

我有一个在我目前编写的代码中很常见的问题，即我想要一个只能存在于范围为 [start, end) 的特定范围内的整数。基本上我希望能够执行如下操作：

cyclic_int ci(4, 8);

ci = 4;
assert(ci == 4);
ci += 3;
assert(ci == 7);
ci += 2;
assert(ci == 5);
assert(ci == 13);

并且这一切都应该返回 true。基本上，该类自动为我应用模数 (%)，并且该整数充当我初始化它的范围内的循环整数。我可以自己实现这个类并重载所有常见的运算符，使其能够与普通整数很好地配合，但它似乎是一个有用的类，可能有人以前做过。

所以我的问题是，是否有一个每个人都使用的公共类，或者我是否正在考虑以错误的方式进行操作，是否有更好的更简单的方法。（我的目标是不必不断考虑应用 % 运算符或任何类似的函数）谢谢。

编辑：我决定也写自己的一个只是为了好玩： http://github.com/robertmassaioli/wrapping_number< /a>

原文

I have a problem that is quite common in the code that I am writing at the moment whereby I want to have an integer that can only exist inside a certain range where the range is [start, end). Basically I want to be able to do something like the following:

cyclic_int ci(4, 8);

ci = 4;
assert(ci == 4);
ci += 3;
assert(ci == 7);
ci += 2;
assert(ci == 5);
assert(ci == 13);

And that should all return true. Basically the class automatically applies modulus (%) for me and the integer acts as a cyclic integer in the range that I init it with. I could implement this class myself and overload all of the common operators to make it work nicely with normal integers but it seems like a useful class that somebody may have made before.

So my question is this, is there a common class like this out there somewhere that everyone uses or am I thinking of doing it the wrong way and is there a better simpler way. (My aim is to not have to constantly think about applying the % operator or any similar function over it) Thanks.

Edit: I decided to write my own one as well just for fun: http://github.com/robertmassaioli/wrapping_number

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吹泡泡o 2024-10-01 21:30:09

使用normalize函数不是更简单吗？

int normalize(int val, int start, int end)
{
    return (val - start) % (end - start) + start;
}


int ci = 4;   
assert(ci == 4);   
ci = normalize(ci + 3, 4, 8);   
assert(ci == 7);   
ci = normalize(ci + 2, 4, 8);   
assert(ci == 5);   
assert(ci == 13);

Is not it easier to use the function normalize?

int normalize(int val, int start, int end)
{
    return (val - start) % (end - start) + start;
}


int ci = 4;   
assert(ci == 4);   
ci = normalize(ci + 3, 4, 8);   
assert(ci == 7);   
ci = normalize(ci + 2, 4, 8);   
assert(ci == 5);   
assert(ci == 13);

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