Calling functions with setTimeout()

Question

Simply put...

why does

setTimeout(\'playNote(\'+currentaudio.id+\', \'+noteTime+\')\', delay);

work perfectly, calling the function

Answer 1

It may help to understand when javascript executes code, and when it waits to execute something:

let foo2 = function foo(bar=baz()){ console.log(bar); return bar()}

• The first thing javascript executes is the function constructor, and creates a function object. You can use either the function keyword syntax or the => syntax, and you get similar (but not identical) results.
• The function just created is then assigned to the variable foo2
• At this point nothing else has been run: no other functions called (neither baz nor bar, no values looked up, etc. However, the syntax has been checked inside the function.
• If you were to pass foo or foo2 to setTimeout then after the timeout, it would call the function, the same as if you did foo(). (notice that no args are passed to foo. This is because setTimeout doesn't by default pass arguments, although it can, but those arguments get evaluated before the timeout expires, not when it expires.)
• After foo is called, default arguments are evaluated. Since we called foo without passing arguments, the default for bar is evaluated. (This would not have happened if we passed an argument)
• While evaluating the default argument for bar, first javascript looks for a variable named baz. If it finds one, it then tries to call it as a function. If that works, it saves the return value to bar.
• Now the main body of the function is evaluated:
• Javascript looks up the variable bar and then calls console.log with the result. This does not call bar. However, if it was instead called as bar(), then bar would run first, and then the return value of bar() would be passed to console.log instead. Notice that javascript gets the values of the arguments to a function it is calling before it calls the function, and even before it looks up the function to see if it exists and is indeed a function.
• Javascript again looks up bar, and then it tries to call it as a function. If that works, the value is returned as the result of foo()

So, function bodies and default arguments are not called immediately, but everything else is. Similarly, if you do a function call (i.e. ()), then that function is executed immediately as well. However, you aren't required to call a function. Leaving off the parentheses will allow you to pass that function around and call it later. The downside of that, though, is that you can't specify the arguments you want the function to be called with. Also, javascript does everything inside the function parentheses before it calls the function or looks up the variable the function is stored in.

Answer 2

Try this.

setTimeout(function() { playNote(currentaudio.id,noteTime) }, delay);

Answer 3

Don't use string-timeouts. It's effective an eval, which is a Bad Thing. It works because it's converting currentaudio.id and noteTime to the string representations of themselves and hiding it in the code. This only works as long as those values have toString()s that generate JavaScript literal syntax that will recreate the value, which is true for Number but not for much else.

setTimeout(playNote(currentaudio.id, noteTime), delay);

that's a function call. playNote is called immediately and the returned result of the function (probably undefined) is passed to setTimeout(), not what you want.

As other answers mention, you can use an inline function expression with a closure to reference currentaudio and noteTime:

setTimeout(function() {
    playNote(currentaudio.id, noteTime);
}, delay);

However, if you're in a loop and currentaudio or noteTime is different each time around the loop, you've got the Closure Loop Problem: the same variable will be referenced in every timeout, so when they're called you'll get the same value each time, the value that was left in the variable when the loop finished earlier.

You can work around this with another closure, taking a copy of the variable's value for each iteration of the loop:

setTimeout(function() {
    return function(currentaudio, noteTime) {
        playNote(currentaudio.id, noteTime);
    };
}(currentaudio, noteTime), delay);

but this is getting a bit ugly now. Better is Function#bind, which will partially-apply a function for you:

setTimeout(playNote.bind(window, currentaudio.id, noteTime), delay);

(window is for setting the value of this inside the function, which is a feature of bind() you don't need here.)

However this is an ECMAScript Fifth Edition feature which not all browsers support yet. So if you want to use it you have to first hack in support, eg.:

// Make ECMA262-5 Function#bind work on older browsers
//
if (!('bind' in Function.prototype)) {
    Function.prototype.bind= function(owner) {
        var that= this;
        if (arguments.length<=1) {
            return function() {
                return that.apply(owner, arguments);
            };
        } else {
            var args= Array.prototype.slice.call(arguments, 1);
            return function() {
                return that.apply(owner, arguments.length===0? args : args.concat(Array.prototype.slice.call(arguments)));
            };
        }
    };
}

Answer 4

The first form that you list works, since it will evaluate a string at the end of delay. Using eval() is generally not a good idea, so you should avoid this.

The second method doesn't work, since you immediately execute a function object with the function call operator (). What ends up happening is that playNote is executed immediately if you use the form playNote(...), so nothing will happen at the end of the delay.

Instead, you have to pass an anonymous function to setTimeout, so the correct form is:

setTimeout(function() { playNote(currentaudio.id,noteTime) }, delay);

Note that you are passing setTimeout an entire function expression, so it will hold on to the anonymous function and only execute it at the end of the delay.

You can also pass setTimeout a reference, since a reference isn't executed immediately, but then you can't pass arguments:

setTimeout(playNote, delay);

---

Note:

For repeated events you can use setInterval() and you can set setInterval() to a variable and use the variable to stop the interval with clearInterval().

You say you use setTimeout() in a for loop. In many situations, it is better to use setTimeout() in a recursive function. This is because in a for loop, the variables used in the setTimeout() will not be the variables as they were when setTimeout() began, but the variables as they are after the delay when the function is fired.

Just use a recursive function to sidestep this entire problem.

Using recursion to deal with variable delay times:

  // Set original delay
var delay = 500;

  // Call the function for the first time, to begin the recursion.
playNote(xxx, yyy);

  // The recursive function
function playNote(theId, theTime)
{
    // Do whatever has to be done
    // ...

    // Have the function call itself again after a delay, if necessary
    //   you can modify the arguments that you use here. As an
    //   example I add 20 to theTime each time. You can also modify
    //   the delay. I add 1/2 a second to the delay each time as an example.
    //   You can use a condition to continue or stop the recursion

    delay += 500;

    if (condition)
    { setTimeout(function() { playNote(theID, theTime + 20) }, delay); }
}

Answer 5

I literally created an account on this site to comment on Peter Ajtai's answer (currently highest voted), only to discover that you require 50 rep (whatever that is) to comment, so I'll do it as an answer since it's probably worth pointing out a couple things.

In his answer, he states the following:

You can also pass setTimeout a reference, since a reference isn't executed immediately, but then you can't pass arguments:

setTimeout(playNote, delay);

This isn't true. After giving setTimeout a function reference and delay amount, any additional arguments are parsed as arguments for the referenced function. The below would be better than wrapping a function call in a function.

setTimeout(playNote, delay, currentaudio.id, noteTime)

Always consult the docs.

That said, as Peter points out, a recursive function would be a good idea if you want to vary the delay between each playNote(), or consider using setInterval() if you want there to be the same delay between each playNote().

Also worth noting that if you want to parse the i of your for loop into a setTimeout(), you need to wrap it in a function, as detailed here.

Answer 6

Because the second one you're telling it to call the playNote function first and then pass the return value from it to setTimeout.