C# 如何仅在协同程序完成后继续该功能？

输出：

时间1-0
时间3-0
时间2-3.0146

我想要的输出是：

时间1-0
时间2-3
时间3-3

位于的文档中似乎有一个与您尝试执行的操作几乎相同的示例：

void Generate()
{
    StartCoroutine(FallDelayCoroutine());
    print("time3- " + Time.time);
}

IEnumerator FallDelayCoroutine()
{     
    print("time1- "+ Time.time);
    yield return new WaitForSeconds(3f);
    print("time2- " + Time.time);
}

关键点似乎是他们的

Start

例程返回IEnumerator，然后使用

yield return Start例程（WaitAndPrint（2.0F））

强制它在继续之前等待该方法

因此，对你来说，等价物是：

IEnumerator Start()
{
    // - After 0 seconds, prints "Starting 0.0"
    // - After 2 seconds, prints "WaitAndPrint 2.0"
    // - After 2 seconds, prints "Done 2.0"
    print("Starting " + Time.time);

    // Start function WaitAndPrint as a coroutine. And wait until it is completed.
    // the same as yield WaitAndPrint(2.0);
    yield return StartCoroutine(WaitAndPrint(2.0F));
    print("Done " + Time.time);
}

// suspend execution for waitTime seconds
IEnumerator WaitAndPrint(float waitTime)
{
    yield return new WaitForSeconds(waitTime);
    print("WaitAndPrint " + Time.time);
}

位于的文档中似乎有一个与您尝试执行的操作几乎相同的示例：

void Generate()
{
    StartCoroutine(FallDelayCoroutine());
    print("time3- " + Time.time);
}

IEnumerator FallDelayCoroutine()
{     
    print("time1- "+ Time.time);
    yield return new WaitForSeconds(3f);
    print("time2- " + Time.time);
}

关键点似乎是他们的

Start

例程返回IEnumerator，然后使用

yield return Start例程（WaitAndPrint（2.0F））

强制它在继续之前等待该方法

因此，对你来说，等价物是：

IEnumerator Start()
{
    // - After 0 seconds, prints "Starting 0.0"
    // - After 2 seconds, prints "WaitAndPrint 2.0"
    // - After 2 seconds, prints "Done 2.0"
    print("Starting " + Time.time);

    // Start function WaitAndPrint as a coroutine. And wait until it is completed.
    // the same as yield WaitAndPrint(2.0);
    yield return StartCoroutine(WaitAndPrint(2.0F));
    print("Done " + Time.time);
}

// suspend execution for waitTime seconds
IEnumerator WaitAndPrint(float waitTime)
{
    yield return new WaitForSeconds(waitTime);
    print("WaitAndPrint " + Time.time);
}

作为使用两个协同路由的替代方案，您还可以添加一个回调，例如

IEnumerator Generate()
{
    yield return StartCoroutine(FallDelayCoroutine());
    print("time3- " + Time.time);
}

并使用lambda表达式调用它

IEnumerator FallDelayCoroutine(Action whenDone)
{     
    print("time1- "+ Time.time);
    yield return new WaitForSeconds(3f);
    print("time2- " + Time.time);

    whenDone?.Invoke();
}

或者用一种方法

void Generate()
{
    StartCoroutine(FallDelayCoroutine(() => {
        print("time3- " + Time.time);
    }));  
}

作为使用两个协同路由的替代方案，您还可以添加一个回调，例如

IEnumerator Generate()
{
    yield return StartCoroutine(FallDelayCoroutine());
    print("time3- " + Time.time);
}

并使用lambda表达式调用它

IEnumerator FallDelayCoroutine(Action whenDone)
{     
    print("time1- "+ Time.time);
    yield return new WaitForSeconds(3f);
    print("time2- " + Time.time);

    whenDone?.Invoke();
}

或者用一种方法

void Generate()
{
    StartCoroutine(FallDelayCoroutine(() => {
        print("time3- " + Time.time);
    }));  
}

---

我想这似乎回答了你的问题（尽管我对团结没有经验）。特别是第二个例子似乎正是您想要的…协同程序的要点是它不会这样做。。。我想这似乎回答了你的问题（尽管我对团结没有经验）。特别是第二个例子似乎正是您想要的…协同程序的要点是它不会这样做。。。