javascript, Javascript bug

Question

Run in console next code

setTimeout(function(){
   console.log(1);
}, 2);
console.log(2);
setTimeout(function(){
   console.log(3);
}, 1);

Why 4 values when has only 3? console

That code does not produce 4 values. Here's proof that it doesn't. Something else is printing out the 20. For future reference, do not include console output or any other text as images. Copy the text and include it in your question. For one, it makes it more searchable. — Mike Cluck
– Mike Cluck, Commented May 31, 2017 at 13:50
You can tell something else is outputting 20 by the little arrow next to it. — dav
– dav, Commented May 31, 2017 at 13:51
You console log shows something called a "return value". The return value of a timeout is the ID for that timeout. — evolutionxbox
– evolutionxbox, Commented May 31, 2017 at 13:51
It is the setTimeout reference, try var t = setTime.. then print console.log(t). Now, why it shows only the first idk. — DontVoteMeDown
– DontVoteMeDown, Commented May 31, 2017 at 13:52

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

You get a return value of setTimeout in the console.

The returned timeoutID is a numeric, non-zero value which identifies the timer created by the call to setTimeout(); this value can be passed to clearTimeout() to cancel the timeout.

It may be helpful to be aware that setTimeout() and setInterval() share the same pool of IDs, and that clearTimeout() and clearInterval() can technically be used interchangeably. For clarity, however, you should try to always match them to avoid confusion when maintaining your code

console.log(setTimeout(function(){
   console.log('#' + 1);
}, 2));
console.log('#' + 2);
console.log(setTimeout(function(){
   console.log('#' + 3);
}, 1));

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 31, 2017 at 13:52

Nina Scholz

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