检查风向是否在指定范围内

发布于 2024-09-10 12:47:25 字数 872 浏览 1 评论 0原文

我使用整数值（枚举）表示风向，范围从 0（北）到 15（北-西北-西）。

我需要检查给定的风向（0 到 15 之间的整数值）是否在一定范围内。我首先指定 WindDirectionFrom 值，顺时针移动到 WindDirectionTo 以指定允许的风向范围。

显然，如果 WindDirectionFrom=0 和 WindDirectionTo=4 （在 N 和 E 方向之间）并且风向为 NE (2)，则计算很简单

int currentWindDirection = 2;
bool inRange = (WindDirectionFrom <= currentWindDirection && currentWindDirection <= WindDirectionTo); 
              //(0 <= 2 && 2 <= 4) simple enough...

但是对于不同的情况来说WindDirectionFrom=15、WindDirectionTo=4 并且风向再次为 NE (2)，计算立即中断...

bool inRange = (WindDirectionFrom <= currentWindDirection && currentWindDirection <= WindDirectionTo); 
              //(15 <= 2 && 2 <= 4) oops :(

我确信这不会太困难，但我对这个有一个真正的心理障碍。

原文

I am representing wind directions using integer values (an Enum) ranging from 0 for North, through to 15 for North-North-West.

I need to check if a given wind direction (integer value between 0 and 15) is within a certain range. I specify my WindDirectionFrom value first moving clockwise to WindDirectionTo to specify the range of allowable wind direction.

Obviously if WindDirectionFrom=0 and WindDirectionTo=4 (between N and E direction) and the wind direction is NE (2) the calculation is simply

int currentWindDirection = 2;
bool inRange = (WindDirectionFrom <= currentWindDirection && currentWindDirection <= WindDirectionTo); 
              //(0 <= 2 && 2 <= 4) simple enough...

However for a different case where say WindDirectionFrom=15, WindDirectionTo=4 and wind direction is NE (2) again, the calculation immediately breaks...

bool inRange = (WindDirectionFrom <= currentWindDirection && currentWindDirection <= WindDirectionTo); 
              //(15 <= 2 && 2 <= 4) oops :(

I'm sure this can't be too difficult, but I'm having a real mental block with this one.

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花伊自在美 2024-09-17 12:47:25

你想要的是模运算。进行模 16 算术，并检查差异是否至少为 14（-2 的模等价物）或最多 2。

如何进行模算术因语言而异。使用 C 或 C++，您会发现 x mod 16 如下：（

int xm = x % 16;
if (xm < 0) xm += 16;

感谢 msw 指出，enum 上的算术经常是不允许的，并且有充分的理由。enum code> 通常表示离散且算术上不相关的对象或条件。）

What you want is modular arithmetic. Do your arithmetic mod 16, and check to see if the difference is from (say) at least 14 (the modular equivalent of -2) or at most 2.

How to do modular arithmetic will vary between languages. With C or C++, you would find x mod 16 as follows:

int xm = x % 16;
if (xm < 0) xm += 16;

(Thanks to msw for pointing out that arithmetic on enums is frequently not allowed, and for good reasons. An enum normally represents objects or conditions that are discrete and not related arithmetically.)

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天暗了我发光 2024-09-17 12:47:25

我会这样做：

int normedDirection( int direction, int norm )
{
   return (NumberOfDirections + currentDirection - norm) % NumberOfDirections;
}

int normed = normedDirection( currentWindDirection, WindDirectionFrom );
bool inRange = (0 <= normed && normed <= normedDirection( WindDirectionTo, WindDirectionFrom );

I would do it like this:

int normedDirection( int direction, int norm )
{
   return (NumberOfDirections + currentDirection - norm) % NumberOfDirections;
}

int normed = normedDirection( currentWindDirection, WindDirectionFrom );
bool inRange = (0 <= normed && normed <= normedDirection( WindDirectionTo, WindDirectionFrom );

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