CODEWITHSUNDEEP
javascriptasynchronous
user6369603
user6369603

Reputation:

How much further does code execute after an asynchronous function is called

Apologies if the title of this question is slightly confusing, I was struggling to find a way to word this.

Basically, when an asynchronous function is called, how much further down the code will be executed if the asynchronous function is called from other functions, objects, or files. For example:

function func1() {
  // async function 1
  // async function 2
}

function func2() {
  // more code
}

func1();
func2();

Let's say we execute func1() and both the async functions inside of it take a very long time to run. Will we continue on to func2() while the content of func1() is still being run, or will we block until func1() is finished before continuing to func2()?

Upvotes: 3

Views: 759

Answers (3)

Maciej Sikora
Maciej Sikora

Reputation: 20132

JavaScript is single threaded, it literally means nothing can be executed in the same time. If your code has asynchronous nature it means that it is executed in uknown future. I said uknown because it depends from the code. For example asynchronous code for ajax call will be executed when the browser gets the response from the server.

That said, the async code is not executed in the declaration moment but in uknown next event loop. Furthermore it not blocks the current execution. In javascript your execution code can not be disturbed. It just runs through all lines.

Going back to the example, if func1 has no sync blocking code then func2 will be executed right away without any delay.

Upvotes: 1

Ajay Gaur
Ajay Gaur

Reputation: 5270

JavaScript is single threaded, non-blocking and asynchronous language. JavaScript has call stack, event loop and a callback queue. Words are taken directly from this video. Javascript works on v8 engine (chrome), spider monkey (firefox) which provide JavaScript call stack and heap. v8 or spider monkey provide call stack to Javascript so whenever a function is called, it is stored in the call stack of the runtime (browser in our case, local if you have node installed). Browser also provide the web apis to the javascript like setTimeOut, XMLHttpRequest and DOM. The pictorial illustration is something like this. (Source is the same video which I've tagged.)

enter image description here

JavaScript is single threaded and it means it can execute one function at a time as it has only one call stack. So whenever an asynchronous code (inside func1)is executing, it is executed through webAPIs provided by the browser. The role of callback queues come here. Whenever the result from the Async code is executed, it is stored in the callback queue and it waits for the stack to get empty (event driven programming). Whenever it sees that the stack is empty, the function from the callback queues will start executing.

In your case, you're calling func1, it kicks in some async code but currently func1 is in stack. If the async code has completed and if it sees the stack as empty, it will executed the async code first and then func2 but if the async code hasn't completed, it will start executing the func2 and the callback queue will wait for the stack to get empty. In this case, the flow will be func1 --> func2 --> async code.

So, it is matter of timing. For async code, if the code has returned and waiting in callback queue, as soon as it sees the stack empty, it will kick in and start executing the callbacks (.then or callback from setTimeOut etc.) I'll recommend you to watch that video for better insights and what an event loop in JS is. If I'm missing out something, anyone please feel free to edit this answer or provide suggestions for edit.

Upvotes: 2

Alexei Levenkov
Alexei Levenkov

Reputation: 100547

Asynchronous callback are top level events (similar to button click events) and can only be started when there is no other scripts executing.

Execution will continue till call stack is empty (including "global user code" which what your sample shows) and script engine itself gets control back - so both func1() and func2() will finish before callbacks are executed.

Note: you can't rely on such behavior for normal code - if "asynchronous" function has both synchronous and asynchronous code branches callback from that function can be executed synchronously without even returning to calling function (this can happen with plain callbacks or any other form including promises).

Upvotes: 0

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