Question

我在这里看到一些东西，它通过将旧pos保存为UNITY来加快速度，并提供一些代码：

void TestVelocity(Vector3& Pos, Vector3 &Result)
{
    Vector3 PreviousPos;

    if (GetTickCount() >= velupd)
    {
        velupd = GetTickCount() + 100//random timer for test;

        Vector3 Diff = Pos - PreviousPos;
        float Len = sqrtf(Diff .x * Diff .x + Diff .y * Diff .y + Diff .z * Diff .z);

        if (Len >= 0.01)
        {
            Result = (Diff / Len);
        }
    }
      PreviousPos = Pos.
}

计算错误。数据仅适用于对象位置（无速度等）。

Answer 1

需要更多详细信息才能为您提供更准确的答案。但是我基本上看到两个问题。您没有初始化 PreviousPos 对象，在我看来您希望它持久存在，对吧？

假设 Vector3 类具有负（-）运算符重载，那么您可以执行此操作，您可以这样做。

void TestVelocity(Vector3& Pos, Vector3 &Result)
{  
static Vector3 PreviousPos; //Initialize here the initial position to zero with your constructor

    if (GetTickCount() >= velupd)
    {
        velupd = GetTickCount() + 100//random timer for test;

        Vector3 Diff = Pos - PreviousPos;
        float Len = sqrtf(Diff .x * Diff .x + Diff .y * Diff .y + Diff .z * Diff .z);

        if (Len >= 0.01)
        {
            Result = (Diff / Len);
        }
    }
  PreviousPos = Pos.
}

另一种解决方案是将前一个位置作为参数，但是正如我所说，很难说我们是否不知道您想要的实现是什么。这些是建议，具体取决于所需的内容。

新旧位置之间的速度

1 个答案: