C # yield and try-finally

Question

If I have a coroutine, will the code in the finally block be called?

public IEnumerator MyCoroutine(int input)
{
  try
  {
    if(input > 10)
    {
      Console.WriteLine("Can't count that high.");
      yield break;
    }
    Console.WriteLine("Counting:");
    for(int i = 0; i < input; i++)
    {
      Console.WriteLine(i.ToString());
      yield return null;
    }
  }
  finally
  {
    Console.WriteLine("Finally!");
  }
}

Answer 1

As long as the iterator / enumerator is correctly located (called IDisposable.Dispose()), then yes:

The control is always passed to the finally block no matter how the try block exits.

http://msdn.microsoft.com/en-us/library/zwc8s4fz.aspx

However, be careful what is Dispose()really called, otherwise you may end up with unintended results. For more information about this phenomenon and some issues to keep in mind, check out this blog post:

- Dispose !

( . , , , , )

:

return yield try-catch. try, try finally.

http://msdn.microsoft.com/en-us/library/9k7k7cf0.aspx

Answer 2

: finally, yield return, , IDisposable.Dispose /, yield return. GetEnumerator() , MoveNext() , yield return finally, Dispose, finally . , , ( ), ( , , - ).

, , vb # , foreach Dispose , , GetEnumerator(), , - Dispose(). , , , .

Answer 3

, , .

yield, finally , IEnumerator, . :

void Main()
{
    var x = MyCoroutine(12);

    //Console.WriteLines will happen when the following
    //statement is executed
    var y = x.MoveNext();
}

Answer 4

Main() .., , , :

YieldReturnAndFinally

@Henk Holterman

:

* IEnumerable
* IEnumerable<T>
* IEnumerator
* IEnumerator<T>