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I was writing a long polling script and ran into a too much recursion error which hung the browser. My goal is to call the same function every 1000ms using setTimeout(). Yes, I could use setInterval() but it is going to be a long polling script and will be waiting for a server response.

I fixed this by removing the () from the function I was calling within the same function.

My script looks like:

function messagePolling(){
    console.log("polled")
    setTimeout(messagePolling(),1000) // <--- removing `()` from the function works as intended 
} 

messagePolling();

What is the logic behind this? messagePolling is a function after all isn't it.

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    messagePolling is a reference to a function. messagePolling() invokes the function. You should know that since you have messagePolling(); in the last line of your code snippet ;) And I'm sure you have passed the return value of a function to other functions before, i.e. foo(bar()). This is will call bar and pass the return value to foo. That's how function invocation works in JavaScript. Commented Sep 24, 2016 at 19:15
  • See e.g. this question or many other duplicates Commented Sep 24, 2016 at 19:29

4 Answers 4

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You're absolutely right - messagePolling is a function. However, messagePolling() is not a function. You can see that right in your console:

// assume messagePolling is a function that doesn't return anything
messagePolling() // -> undefined

So, when you do this:

setTimeout(messagePolling(), 1000)

You're really doing this:

setTimeout(undefined, 1000)

But when you do this:

setTimeout(messagePolling, 1000)

You're actually passing the function to setTimeout. Then setTimeout will know to run the function you passed - messagePolling - later on. It won't work if it decides to call undefined (the result of messagePolling()) later, right?

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Written as setTimeout(messagePolling(),1000) the function is executed immediately and a setTimeout is set to call undefined (the value returned by your function) after one second. (this should actually throw an error if ran inside Node.js, as undefined is not a valid function)

Written as setTimeout(messagePolling,1000) the setTimeout is set to call your function after one second.

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When you type messagePolling you are passing the function to setTimeout as a parameter. This is the standard way to use setTimeout.

When you type messagePolling() you are executing the function and passing the return value to setTimeout

That being said, this code looks odd to me. This function just runs itself. It's going to keep running itself indefinitely if you do this.

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Anywhere a function name contains "()" it is executed immediately except when it is wrapped in quotes i.e is a string.

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In which case, it is not a function but a string. That string will only be then used to execute the function if it gets passed through eval which is exactly what setTimeout does if you do pass it a string.
E.g var func = "myFunction()". setTimout("myFunction()",1000) will still be executed. but not instantly. In short I meant function name as a string or variable can not run on its own when just placed in your code without using other functions.
It isn't a function. That's the point your func variable contains a string. It just happens to be code that you can execute, however, you make it sound like it's inherently a delayed execution. It isn't. If you have setTimeout("a = b + c; console.log(a)") that will also work, but only because it gets passed through eval. If you pass the same string through something that does not do eval it will never work.
Yes it is not a function but a string containing function name and "()" which can be executed as a function so eval("myFunction()")==myFunction()
I guess what is strange about this answer is that you can say this about any other kind of expression: "Anywhere you have 1 + 1, the numbers will be added and you get 2, except when it is wrapped in quotes, i.e. is a string.". So, how is having code inside a string relevant here at all? Doesn't really provide any value IMO.
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