这是检测负数的最佳方法吗？ [英] Is this the best way to detect negative numbers?

问题描述

这是我的代码，我真的不能多说了。我的意思是，它工作正常，但它真的是最有效的方式吗？它看起来很脏。 D：

Here is my code, I can''t really say more to it. I mean, it works fine, but is it really the most effiecient way? It looks so dirt. D:

// -- Distance.
/// <summary>
/// Returns the Pixel distance between two points.
/// </summary>
/// <param name="Entity"></param>
/// <returns></returns>
public int GetDistanceFromEntity(Entity Entity)
{
    /* Return distance. */
    int OurEntityX = this.PositionX.ToString().StartsWith("-") ? -this.PositionX : this.PositionX;
    int OurEntityY = this.PositionY.ToString().StartsWith("-") ? -this.PositionY : this.PositionY;

    int OtherEntityX = Entity.PositionX.ToString().StartsWith("-") ? -Entity.PositionX : Entity.PositionX;
    int OtherEntityY = Entity.PositionY.ToString().StartsWith("-") ? -Entity.PositionY : Entity.PositionY;

    return (int)Math.Sqrt(Math.Pow(OtherEntityX - OurEntityX, 2) + Math.Pow(OtherEntityY - OurEntityY, 2));
}

推荐答案

此代码无法返回两个任意点之间的正确距离。

如果 this.PositionX == -2 和 Entity.PositionX == +2 ，之间的区别它们是4，但因为你 有效地 在减法之前取每个的绝对值，你得到的差异将是0 （零！）

对于此计算，只需使用 .PositionX 和 .PositionY 值按原样。

数学将全部解决！



请参阅其他解决方案以获得原始问题的答案...

This code cannot possibly return the correct distance between two arbitrary points.
If this.PositionX == -2 and Entity.PositionX == +2, the difference between them is 4, but because you are effectively taking the absolute value of each before the subtraction, the difference you get will be 0 (zero!)
For this calculation, just use the .PositionX and .PositionY values AS IS.
The math will all work out!

See the other solutions for answers to the original question...

我假设您的值为 int 。

有一个名为的函数Math.Abs​​（...）。





你的功能坏了。坐标可能一般是积极的也可以是消极的（这一点可能存在于四个象限中的任何一个[< a href =http://en.wikipedia.org/wiki/Quadrant_(plane_geometry）target =_ blanktitle =New Window> ^ ]）。如果您在之前取坐标的绝对值，则计算坐标距离会导致错误的结果。按原样取值。

I assume your values are int.
There is a function called Math.Abs(...).


Your function is broken. Coordinates may be in general positive as well as negative (a point may lay in any of the four quadrants[^]). If you take the absolute value of the coordinates before you calculate the coodinate distance results in wrong results. Take the values as they are.

public int GetDistanceFromEntity(Entity Entity)
{
   int deltaX = Entity.PositionX-PositionX;
   int deltaY = Entity.PositionY-PositionY;
   return (int)Math.Sqrt(deltaX*deltaX + deltaY*deltaY);
}

您可以在所有象限和两个方向上使用积分进行测试：

You can test it with points in all quadrants and in both directions:

var q1=new Entity(1, 2);
var q2=new Entity(-3, 2);
var q3=new Entity(-3, -1);
var q4=new Entity(4, -1);
var p0=new Entity(0, 0);

Assert.AreEqual(0, q1.GetDistanceFromEntity(q1), "identiy should have distance 0");
Assert.AreEqual(0, q2.GetDistanceFromEntity(q2), "identiy should have distance 0");
Assert.AreEqual(0, q3.GetDistanceFromEntity(q3), "identiy should have distance 0");
Assert.AreEqual(0, q4.GetDistanceFromEntity(q4), "identiy should have distance 0");
Assert.AreEqual(0, p0.GetDistanceFromEntity(p0), "identiy should have distance 0");

Assert.AreEqual(q2.GetDistanceFromEntity(q1), q1.GetDistanceFromEntity(q2), "distance should be invariant");
Assert.AreEqual(q3.GetDistanceFromEntity(q1), q1.GetDistanceFromEntity(q3), "distance should be invariant");
Assert.AreEqual(q4.GetDistanceFromEntity(q1), q1.GetDistanceFromEntity(q4), "distance should be invariant");
Assert.AreEqual(p0.GetDistanceFromEntity(q1), q1.GetDistanceFromEntity(p0), "distance should be invariant");

Assert.AreEqual(4, q1.GetDistanceFromEntity(q2), "distance q1-q2 mismatch");
Assert.AreEqual(5, q1.GetDistanceFromEntity(q3), "distance q1-q3 mismatch");
...

[/ EDIT]



干杯

Andi

[/EDIT]

Cheers
Andi

我认为你要找的是 Math.Abs​​方法 [ ^ ]总是得到正数。

I think what you are looking for is the Math.Abs Method[^] to always get the positive number.

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