在继续使用C#Unity功能之前,请等待协程完成

Question

我正在努力让一个单位在Unity2d中通过网格移动.我让运动顺利运转.我希望函数MovePlayer等到协程完成后再继续,所以程序将等到玩家完成移动后再发出更多订单.

这是我的代码:public class Player:MonoBehaviour {

public Vector3 position;
private Vector3 targetPosition;

private float speed;

void Awake ()
{
    speed = 2.0f;
    position = gameObject.transform.position;
    targetPosition = position;
    GameManager.instance.AddPlayerToList(this);                     //Register this player with our instance of GameManager by adding it to a list of Player objects. 
}

//Function that moves the player, takes a list of nodes as path
public void MovePlayer(List<Node> path)
{
    StartCoroutine(SmoothMovement(path));
    //Next step should wait until SmoothMovement is finished
}

private IEnumerator SmoothMovement(List<Node> path)
{
    float step = speed * Time.deltaTime;

    for (int i = 0; i < path.Count; i++)
    {
        targetPosition = new Vector3(path[i].coordinatesX, path[i].coordinatesY, 0f);

        float sqrRemainingDistance = (transform.position - targetPosition).sqrMagnitude;

        while (sqrRemainingDistance > float.Epsilon)
        {
            transform.position = Vector3.MoveTowards(transform.position, targetPosition, step);
            sqrRemainingDistance = (transform.position - targetPosition).sqrMagnitude;
            yield return null;
        }

        position = transform.position;
    }

}

Answer 1

您不能等待主线程中的函数中的协程,否则您的游戏将冻结,直到您的函数结束.

为什么不在你的协程结束时给你的下一步打电话？

private IEnumerator SmoothMovement(List<Node> path)
{
    float step = speed * Time.deltaTime;

    for (int i = 0; i < path.Count; i++)
    {
        targetPosition = new Vector3(path[i].coordinatesX, path[i].coordinatesY, 0f);

        float sqrRemainingDistance = (transform.position - targetPosition).sqrMagnitude;

        while (sqrRemainingDistance > float.Epsilon)
        {
            transform.position = Vector3.MoveTowards(transform.position, targetPosition, step);
            sqrRemainingDistance = (transform.position - targetPosition).sqrMagnitude;
            yield return null;
        }

        position = transform.position;
    }
    //Next Step
}