Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
192 views
in Technique[技术] by (71.8m points)

javascript - Call asynchronous function recursively

I have an asynchronous function that I want to call multiple times in a row. The problem is that "multiple" can be a few hundred thousand or millions...

The obvious way is to call the same function from the callback like that:

function foo()
{
    asyncBar(foo);
}

Of course some logic is involved to stop the recursion. The question is whether the stack is filling with calls and may cause stackoverflow at some point?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

The question is whether the stack is filling with calls and may cause stackoverflow at some point?

No. If asyncBar() calls the callback it is passed asynchronously, then there is no stack build-up.

In your code:

function foo() {
    asyncBar(foo);
}

here is what is happening, step-by-step:

  1. First foo() is called.
  2. This then calls asyncBar(foo).
  3. Because asyncBar is asynchronous, that means it starts an asynchronous operation (let's suppose it is an http GET, but any async operation will do). That asynchronous operation is initiated, but then asyncBar() immediately returns.
  4. That initial call to foo() returns and the stack is completely unwound. There is no foo() on the stack any more.
  5. Any code after the call to foo() continues to run until it is done and returns back to the event loop.
  6. Meanwhile the asynchronous operation finishes some time in the future. That places a call to your callback in the event queue.
  7. When the JS engine is done executing other Javascript (which means the stack is completely empty), it pulls that event out of the event queue and calls the callback.
  8. In this case, the callback function is foo so it calls that function and starts the cycle all over again, right back to step 2.

There is no stack build-up. The key is that asynchronous callbacks are called sometime later, after the current stack has finished, unwound and returned back to the system.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...