本文介绍了如何停止合规?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

当两人合作程序正在运行,你怎么停止第一次合作例行?



  GLOBALS.stableTime = 5 ; 

的IEnumerator StableWaittingTime()
{
收益回报新WaitForSeconds(1F);
如果(GLOBALS.stableTime == 0){
GameManager.instance.LevelFaildMethod();
}其他{
GameManager.instance.stableWaittingTime.text = GLOBALS.stableTime.ToString();
GLOBALS.stableTime--;
StartCoroutine(StableWaittingTime);
}
}


解决方案

@ Imapler 答案几乎是你所需要的。我只想补充一点, StopCoroutine 的方法 MonoBehaviour 是超载,有3个类型的参数,所以可能停止同名的很多协同程序。
。对于在这里您需要,只需使用产量突破;像这样的:

 无效的start()
{
StartCoroutine(StableWaittingTime());
}

的IEnumerator StableWaittingTime()
{
收益回报新WaitForSeconds(1F);
如果(假)
{
//做些什么
}
,否则
{
//做些什么
StartCoroutine( StableWaittingTime());
产量突破;
}
}


When two co-routines are running, how do you stop the first co-routine?

GLOBALS.stableTime = 5;

IEnumerator StableWaittingTime ()
{
        yield return new WaitForSeconds (1f);
        if (GLOBALS.stableTime == 0) {
                GameManager.instance.LevelFaildMethod ();
        } else {
                GameManager.instance.stableWaittingTime.text = GLOBALS.stableTime.ToString ();
                GLOBALS.stableTime--;
                StartCoroutine ("StableWaittingTime");
        }
}
解决方案

@Imapler answer is almost all you need. I would just add that StopCoroutine method of MonoBehaviour is overloaded and has 3 types of parameters, so it is possible to stop many coroutines of same name.For your need here, just use yield break; like this:

void Start ()
{
    StartCoroutine (StableWaittingTime ());
}

IEnumerator StableWaittingTime ()
{
    yield return new WaitForSeconds (1f);
    if (false)
    {
        // do something
    }
    else
    {
        // do something
        StartCoroutine (StableWaittingTime ());
        yield break;
    }
}

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09-03 02:57