I am not so into JavaScript and I have the following doubt related the setTimeout() method.
So into a test script I have:
function simpleMessage() {
alert("This is just an alert box");
}
// settimeout is in milliseconds:
setTimeout(simpleMessage, 5000);
So when I perform the page, after 5 second the simpleMessage() function is performed and it is shown the alert popup.
I understand that when I do:
setTimeout(simpleMessage, 5000);
it means that the simpleMessage() function have to be performed after 5 second after the timer settings but why it is used simpleMessage and not simpleMessage() for the function invocation?
simpleMessage is a reference to a function whereas simpleMessage() executes the function. setTimeout needs a function reference to call a later time.
To perhaps make things a little more obvious, you could have written your function declaration as
// define my function (but don't execute it)
var myFunction = function() {
alert('SOUND THE ALARMS!');
};
// start a timer that will execute the given function after the given
// period of time
setTimeout(myFunction, 5000);
See setTimeout documentation.
The first argument to setTimeout is the function to be executed. The identifier simpleMessage refers to the function you want setTimeout to execute, so that's what you supply as an argument to setTimeout.
If you did setTimeout(simpleMessage(), 5000);, you would execute simpleMessage immediately and then setTimeout would get the return value as its first argument. This is comparable to:
var value = simpleMessage();
setTimeout(value, 5000);
This doesn't make sense; it is the same as setTimeout(simpleMessage(), 5000);.
Consider also a higher-order function that returns a function:
function funcFacotry() {
return function() { alert("this is just an alert box."); }
}
var simpleMessage = funcFactory();
setTimeout(simpleMessage, 5000);
In this case, this actually does make sense, because the return value of funcFactory is actually a function itself.
setTimeout in javascript executes the function after a specific amount of time set as the second parameter, while if you use with setInterval it will execute in intervals, without to consider if the function is get executed or not (this will lead to chunkiness for example if you ar using for animation).
As a metter of second question: if you are using the function with parentheses, this is a method invocation, while using without parentheses is a reference to a specific method.
Because you need to pass a reference to setTimeout of the function you want to invoke after the 5s.
This:
setTimeout(simpleMessage(), 5000);
would execute the simpleMessage function at the same time you're calling the setTimeout function.
Related
I am new to js.
I am trying to make a function that only calls input function f every 50 milliseconds.
wrote the code in the fiddle using setTimeout.
but isn't doing exactly what I wanted it to do
can you tell me how to fix it.
providing code below
https://jsfiddle.net/1pqznbgt/
function callSeconds(f){
setTimeout(function(){
alert("testing");
}, 500)
}
You are looking for setInterval instead of setTimeout (setTimeout will execute once, setInterval will execute at an interval)
The 500 in your sample means 500ms rather than 50ms
You will want to call f rather than your anonymous function, so:
What you are looking for is something like:
function callSeconds(f){
setInterval(f, 50);
}
The problem I see is that every time you call an alert() you are stopping execution of JavaScript and each browser can implement the alert differently . setTimeout() only executes once so if you call the function it will wait 500ms and issue an alert if you want a piece of code to operate continually you must use the setInterval() function but you still have the problem of how the browser chooses to handle the alert() function, normally an alert pauses execution of the current code, what effect that has on a timed execution function I do not know. I would try execution on something that does not pause or stop execution of the script. I also notice that in the code segment you have not invoked the parent function that would call the setTimeout function.
You can do one of 2 things here.
First, don't use setTimeout.
Instead use setInterval (f,50).
Or recursively call the function as so:
function foo (f){
f();
setTimeout (foo,50);
}
I need to run a javascript function each 10 seconds.
I understand the syntax must work like follow but I am not getting any success:
function funcName() {
alert("test");
}
var func = funcName();
var run = setInterval("func",10000)
But this isn't working. Any help?
A lot of other answers are focusing on a pattern that does work, but their explanations aren't really very thorough as to why your current code doesn't work.
Your code, for reference:
function funcName() {
alert("test");
}
var func = funcName();
var run = setInterval("func",10000)
Let's break this up into chunks. Your function funcName is fine. Note that when you call funcName (in other words, you run it) you will be alerting "test". But notice that funcName() -- the parentheses mean to "call" or "run" the function -- doesn't actually return a value. When a function doesn't have a return value, it defaults to a value known as undefined.
When you call a function, you append its argument list to the end in parentheses. When you don't have any arguments to pass the function, you just add empty parentheses, like funcName(). But when you want to refer to the function itself, and not call it, you don't need the parentheses because the parentheses indicate to run it.
So, when you say:
var func = funcName();
You are actually declaring a variable func that has a value of funcName(). But notice the parentheses. funcName() is actually the return value of funcName. As I said above, since funcName doesn't actually return any value, it defaults to undefined. So, in other words, your variable func actually will have the value undefined.
Then you have this line:
var run = setInterval("func",10000)
The function setInterval takes two arguments. The first is the function to be ran every so often, and the second is the number of milliseconds between each time the function is ran.
However, the first argument really should be a function, not a string. If it is a string, then the JavaScript engine will use eval on that string instead. So, in other words, your setInterval is running the following JavaScript code:
func
// 10 seconds later....
func
// and so on
However, func is just a variable (with the value undefined, but that's sort of irrelevant). So every ten seconds, the JS engine evaluates the variable func and returns undefined. But this doesn't really do anything. I mean, it technically is being evaluated every 10 seconds, but you're not going to see any effects from that.
The solution is to give setInterval a function to run instead of a string. So, in this case:
var run = setInterval(funcName, 10000);
Notice that I didn't give it func. This is because func is not a function in your code; it's the value undefined, because you assigned it funcName(). Like I said above, funcName() will call the function funcName and return the return value of the function. Since funcName doesn't return anything, this defaults to undefined. I know I've said that several times now, but it really is a very important concept: when you see funcName(), you should think "the return value of funcName". When you want to refer to a function itself, like a separate entity, you should leave off the parentheses so you don't call it: funcName.
So, another solution for your code would be:
var func = funcName;
var run = setInterval(func, 10000);
However, that's a bit redundant: why use func instead of funcName?
Or you can stay as true as possible to the original code by modifying two bits:
var func = funcName;
var run = setInterval("func()", 10000);
In this case, the JS engine will evaluate func() every ten seconds. In other words, it will alert "test" every ten seconds. However, as the famous phrase goes, eval is evil, so you should try to avoid it whenever possible.
Another twist on this code is to use an anonymous function. In other words, a function that doesn't have a name -- you just drop it in the code because you don't care what it's called.
setInterval(function () {
alert("test");
}, 10000);
In this case, since I don't care what the function is called, I just leave a generic, unnamed (anonymous) function there.
Change setInterval("func",10000) to either setInterval(funcName, 10000) or setInterval("funcName()",10000). The former is the recommended method.
That's because you should pass a function, not a string:
function funcName() {
alert("test");
}
setInterval(funcName, 10000);
Your code has two problems:
var func = funcName(); calls the function immediately and assigns the return value.
Just "func" is invalid even if you use the bad and deprecated eval-like syntax of setInterval. It would be setInterval("func()", 10000) to call the function eval-like.
Try this:
function funcName() {
alert("test");
}
var run = setInterval(funcName, 10000)
Btw this didn't work
setInterval(function () {
alert("test");
}, 10000);
I had to give a name (whatever you like) to the anonymous function:
setInterval(function alertNotification () {
alert("test");
}, 10000);
I am beginner in javascript.
There is strange thing in javascript and I feel stupid
//First statement
var myVar = "Hello";
function hello() {
document.getElementById("demo").innerHTML = myVar;
}
//second statement
var myVar = setInterval(myTimer, 1000);
function myTimer() {
document.getElementById("demo").innerHTML = new Date().toLocaleTimeString();
}
Why the second function work without invoked it ?
Unlike the first ? This caused a lot of problems to me !
Why the second function work without invoked it ? Unlike the first ?
You are passing the function myTimer to setInterval. setInterval calls the function every 1000ms. So while it is not you who directly calls the function, it is still called (by setInterval). setInterval's whole purpose is to call the function that you pass to it as argument.
In contrast, you are not doing anything with hello. You are neither calling it directly, nor are you passing it to any other function that could call it.
From what I can gather from the comments, you seem to be confused as to why the myTimer() function works. Here's a brief explanation:
On this line
var myVar = setInterval(myTimer, 1000);
you are calling the setInterval() function. That function takes 2 parameters, which you have defined. The first one is a function; the code to be executed. The second one is the delay between each execution of said function.
On the next line, you have declared the variable myTimer to be a function which is executed with the setInterval.
Have a look at the MDN documentation for details. Specifically, it says:
var intervalID = window.setInterval(func, delay)
The parameters are defined as:
func: A function to be executed every delay milliseconds.
and
delay: The time, in milliseconds (thousandths of a second), the timer should delay in between executions of the specified function or code.
In JavaScript, the setTimeout function requires code to be enclosed in a function.
Examples of invalid timeouts:
window.setTimeout(console.log('PING'),3000)
function ping(){console.log('PING')};window.setTimeout(ping(),3000)
Example of valid timeouts:
window.setTimeout(function(){console.log('PING')},3000)
function ping(){console.log('PING')};window.setTimeout(function(){ping()},3000)
Now my question: why? I understand why normal code might need to be enclosed in a function, but why is it ALWAYS necessary to enclose code in function(){}?
It doesn't always require an anonymous function. You can also pass a function reference as the first argument, for example, let's assume you have a function called log defined. You can validly write:
function log()
{
console.log( 'PING' );
}
window.setTimeout( log, 200 );
Notice that we don't pass the parentheses with the first argument here.
However, you're not able to pass parameters directly to log() in this instance, so it's necessary to wrap the function call inside an anonymous function.
The code is required to be enclosed in a function because, the setTimeout function does not execute individual lines of code. It takes two, arguments - the first argument is a callback function, and the second argument is the time in milliseconds. The callback function will be called by the setTimeout function internally after the specified amount of time passes.
In the example you gave
window.setTimeout(function(){console.log('PING')},3000)
you pass an anonymous function which would be called after 3000 milliseconds or 3 seconds.
Basically because setTimeout is an asynchronous operation and you need to give what to do next once the timeout is done (i.e. you give a callback function).
Unless JavaScript runtime could block the UI thread and resume execution once the setTimeout ends, you need to provide some executable code to perform an action once the asynchronous operation has been completed:
setTimeout(function() {
// Do stuff after 1 second
}, 1000);
In the other hand, what would be the point of giving a function return value as argument? Check this wrong code to be more illustrative:
// You do some action even BEFORE the timeout operation has been completed?
setTimeout(doStuff(), 1000);
Further reading
Callbacks in Wikipedia
It doesn't need to be enclosed in function(){} but it does need to be a parameter of type Function.
In the case of window.setTimeout(console.log('PING'),3000), what would happen is that console.log() would immediately be executed and the return value (which is undefined) would be passed to the setTimeout function. This code isn't passing a function as a parameter, it's passing the return value of a function as a parameter. Essentially, it's just a shorter way of writing this:
var retVal = console.log('PING'); // retVal === undefined
window.setTimeout(retVal,3000);
That's not special to setTimeout. console.log without () is a function, but console.log() means to invoke that function.
There are other methods to pass a function in to setTimeout, but the anonymous function is typically the cleanest.
Technically, this would work, too:
window.setTimeout(console.log,3000)
but it would not allow you to specify a parameter, making it rather useless here. This could be avoided by binding parameters:
window.setTimeout(console.log.bind(null,'PING'),3000)
In this case, bind is a function which is being invoked (as you can see by the fact it has parameters supplied), so just as before bind is immediately executed. However, bind is a function whose return value is itself a function, and it's that returned function that is passed to setTimeout and called three seconds later. This technique (which you will see used with bind, call, and apply) is called partial application which really just means that you transform a function taking some parameters into a function taking fewer parameters (in this case, zero parameters) by specifying the parameters now but executing the function later.
Because setTimeout is a function that takes two argument - a function, and a number of seconds before that function executes.
https://developer.mozilla.org/en-US/docs/Web/API/WindowTimers/setTimeout
Parameters:
func: A function to be executed after the timer expires.
code: An
optional syntax allows you to include a string instead of a function,
which is compiled and executed when the timer expires. This syntax is
not recommended for the same reasons that make using eval() a security
risk.
delay: Optional The time, in milliseconds (thousandths of a
second), the timer should wait before the specified function or code
is executed. If this parameter is omitted, a value of 0 is used. Note
that the actual delay may be longer; see Reasons for delays longer
than specified below.
param1, ..., paramN Optional Additional
parameters which are passed through to the function specified by func
once the timer expires.
Apparently YOU can just use code, as per the second argument above, but I have never seen it done.
Simply put...
why does
setTimeout('playNote('+currentaudio.id+', '+noteTime+')', delay);
work perfectly, calling the function after the the specified delay, but
setTimeout(playNote(currentaudio.id,noteTime), delay);
calls the function playNote all at the same time?
(these setTimeout()s are in a for loop)
or, if my explanation is too hard to read, what is the difference between the two functions?
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); }
}
Try this.
setTimeout(function() { playNote(currentaudio.id,noteTime) }, delay);
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)));
};
}
};
}
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.
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.
Because the second one you're telling it to call the playNote function first and then pass the return value from it to setTimeout.