我目前在我的程序以下行。我有另外两个整数变量， X 和是。

I have currently the following line in my program. I have two other whole number variables, x and y.

我想看看这个新点（X，Y）是在这条线。我一直在寻找以下主题：

I wish to see if this new point(x, y) is on this line. I have been looking at the following thread:

Given开始和结束点，以及一个距离，计算一个点沿着线的

我已经出来了以下内容：

I've come up with the following:

if(x >= x1 && x <= x2 && (y >= y1 && y <= y2 || y <= y1 && y >= y2))
{
    float vx = x2 - x1;
    float vy = y2 - y1;
    float mag = sqrt(vx*vx + vy*vy);
    // need to get the unit vector (direction)
    float dvx = vx/mag; // this would be the unit vector (direction) x for the line
    float dvy = vy/mag; // this would be the unit vector (direction) y for the line

    float vcx = x - x1;
    float vcy = y - y1;
    float magc = sqrt(vcx*vcx + vcy*vcy);
    // need to get the unit vector (direction)
    float dvcx = vcx/magc; // this would be the unit vector (direction) x for the point
    float dvcy = vcy/magc; // this would be the unit vector (direction) y for the point

    // I was thinking of comparing the direction of the two vectors, if they are the same then the point must lie on the line?
    if(dvcx == dvx && dvcy == dvy)
    {
        // the point is on the line!
    }
}

这似乎并不奏效，或者是这种想法乱作一团？

It doesn't seem to be working, or is this idea whack?

推荐答案

浮点数有一个有限的precision，所以你会得到计算舍入误差，结果是值应该在数学上是平等的意志最终略有不同。

Floating point numbers have a limited precision, so you'll get rounding errors from the calculations, with the result that values that should mathematically be equal will end up slightly different.

您将需要一个小容忍误差比较：

You'll need to compare with a small tolerance for error:

if (std::abs(dvcx-dvx) < tolerance && std::abs(dvcy-dvy) < tolerance)
{
    // the point is (more or less) on the line!
}

最艰难的部分是选择宽容。如果你不能接受任何错误，那么你需要使用的东西比固定precision浮点值等 - 可能是整数，计算重新安排，以避免分裂和其他不准确的操作。

The hard part is choosing that tolerance. If you can't accept any errors, then you'll need to use something other than fixed-precision floating point values - perhaps integers, with the calculations rearranged to avoid division and other inexact operations.

在任何情况下，你可以做到这一点更简单地说，如果没有一个像平方根任何东西。你想看看这两个矢量平行;它们是如果向量积为零或，等价地，如果它们具有相等的切线。所以，你只需要

In any case, you can do this more simply, without anything like a square root. You want to find out if the two vectors are parallel; they are if the vector product is zero or, equivalently, if they have equal tangents. So you just need

if (vx * vcy == vy * vcx)  // might still need a tolerance for floating-point
{
    // the point is on the line!
}

如果你的输入都是整数，足够小，乘法不会溢出，那么就没有必要进行浮点运算都没有。

If your inputs are integers, small enough that the multiplication won't overflow, then there's no need for floating-point arithmetic at all.