在继续使用函数C＃Unity之前等待协程完成

Question

我正在努力使单元在Unity2d中通过网格移动。我得到了运动没有问题的工作。 我希望MovePlayer函数等到协程完成后再继续下去，所以程序会一直等到玩家完成移动后再发出更多的订单。

这里是我的代码： 公共类球员：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

你不能等待在主线程功能的协同程序，否则你的游戏将冻结，直到你的函数结束。

为什么不在协同程序结束时调用下一步？

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 
}