Priority of function execution in javascript - Stack Overflow

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I don't understand why a second function call ( after a function body ) has a priority over the one inside of a body ?

function a(){
  var num = 5;
  console.log( ++num );
  setTimeout( a, 100 );
};
setTimeout(a,2000)

I don't understand why a second function call ( after a function body ) has a priority over the one inside of a body ?

function a(){
  var num = 5;
  console.log( ++num );
  setTimeout( a, 100 );
};
setTimeout(a,2000)

• javascript
• function
• settimeout
• order-of-execution

Share Improve this question Follow edited Feb 7, 2019 at 3:05 Joshua 43.3k99 gold badges7878 silver badges149149 bronze badges asked Jun 7, 2012 at 12:49 Miroslav TrninicMiroslav Trninic 3,45144 gold badges3131 silver badges5555 bronze badges 4

• 2 What behaviour do you expect? – Sirko Commented Jun 7, 2012 at 12:52
• Were you supposed to write "function a() {...}();" (note the final parentheses)? Your code doesn't call the function before the final line. – tsiki Commented Jun 7, 2012 at 12:54
• I just want to know why one call is waiting for the other ? Then I'll be able to predict behaviour. – Miroslav Trninic Commented Jun 7, 2012 at 12:55
• Are you confused by the fact that num never seems to change? If so, it's because you're reinitializaing a new variable every time a is called. – user1106925 Commented Jun 7, 2012 at 13:00

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3 Answers 3

Sorted by: Reset to default 7

In chronological order:

• you are defining function a without calling it

• you are scheduling a to be invoked after two seconds: setTimeout(a,2000)

• it is called

• when it is called, it schedules itself for invocation after 100 milliseconds

Your code basically sleeps for 2 seconds and then executes a with 100 millisecond pauses^[*].

However judging by your context you are asking what is the priority in the following situation:

setTimeout(a, 2000);
setTimeout(b, 100);

Well, most likely b will be called first (assuming there is no unpredictable pause between first and second line, e.g. due to overall OS performance problem).

If you use the same timeouts:

setTimeout(a, 100);
setTimeout(b, 100);

a will most likely be called first. However I don't think this is guaranteed and depends on the JS engine (whether it uses a strict FIFO list for uping events, what is the internal clock resolution, etc.)

^{[*] You can achieve similar behaviour by using setInterval() once.}

The function a isn't called, just defined. The piece of code that is actually run is the definition of a, then setTimeout(a,2000) is called.

I think

function a () {
    var num = 5;
    console.log( ++num );
    setTimeout( a, 100 ); 
};

is a function body and after this we are calling. I do not think it is a hierarchy problem.

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