In working with some javascript I have come across behavior which is puzzling to me and I am looking for some documentation which explains what is going on.
The issue seems to be that depending upon how the functions are passed, they might be executed when I would think they are simply parameters.
Here is a simple example:
function f1()
{
alert('f1');
}
function f2()
{
alert('f2');
}
function f3(a, b)
{
alert('f3');
}
Two versions of calling the above:
f3(f1, f2); // shows 1 alert
f3(f1(), f2()); // shows 3 alerts
The first call of f3 above results in the the f1 and f2 functions not being called and you get one alert for "f3". The second call of f3 functions being called, so you get three alerts for "f1", "f2", and "f3".
In actual code I will be using the first version so that f3 can decide if it wants to call f1 and f2. It took me a bit of fiddling to get it right.
I did not expect this behavior and thought that either with or without the parenthesis the f1 and f2 would not be called just by having them be parameters.
Again, I am looking for some documentation that explains how this works.
In your second line:
f3(f1(), f2());
Your are not passing f1 and f2, you are calling them, and passing their (undefined) return values to f3.
Simply put, adding () to a function, even when passing as a parameter will invoke the function immediately. Example of setting a timeout:
function doStuff() {
alert("ok");
}
setTimeout(doStuff(), 300);
You will be alerted ok immediately because the function is invoked immediately. Setting the timeout with:
setTimeout(doStuff, 300);
Will only alert now when the timeout is called, because you are passing the function in correctly as a parameter.
You will commonly see functions passed in as parameters like this in AJAX callback functions.
The documentation for setTimeout has a great section on callback functions: https://developer.mozilla.org/en-US/docs/Web/API/window.setTimeout
(Yes I realize setTimeout was not used in the question, but I felt this example helped answer the question nicely)
You logic is puzzling. You are passing parameters you never user but let's roll
with the punches.
Your question regards the number of alarms;
You are passing parameters to function that never uses them and you invoke it.
so what actually happens is:
function f3(a, b)
{
console.log("argument 1: " + arguments[0]); //Access the arguments fed to the function
console.log("argument 2: " + arguments[1]);
alert('f3');
}
*
f3(f1, f2);
Pops alert "f3"*
and logs:
argument 1: function f1()
{
alert('f1');
}
argument 2: function f2()
{
alert('f2');
}
f3(f1(),f2())
pops all the alerts
and logs:
argument 1 undefined
argument 2 undefined
When a function has parameters its very clear that parenthesis are used to immediately call it
function sum(a, b){
return a + b;
}
console.log( sum(1, 2) ); //prints 3
The f() version is just a special case of this where your parameter list is empty.
Related
For example:
const factory = {
myFunc(str1) {
console.log(str1)
return (comp) => {
return comp;
}
}
}
console.log(factory.myFunc("foo")("bar"));
The myFunc has four parentheses: factory.myFunc("foo")("bar").
How do you call such a function?
Its called function currying.
Actually read it like factory.myFunc("foo") is returning a function (say x) and calling that function immediately with "bar" argument (like x("bar")).
I don't know about the name, but what you are describing is actually two seperate functions. myFunc simply returns another function.
The line factory.myFunc("foo")("bar") first runs the myFunc funtion with foo as a parameter, this returns another function, which is then immediately run with the parameter bar.
It takes 2 parentheses & 1 argument to call myFunc, you need other 2 parentheses & 1 argument to call function it returns.
This sequence of calls is commonly called chaining of function calls. Sometimes, this sequence of actions is also referred to as pipe.
There's also such a term as currying but I'd say it rather describes a technique of declaring functions in a special way that enables you to pass arguments 1 by 1.
In this case, however, I wouldn't say it's currying since myFunc does something that has nothing to do with what returned function does while in currying (as it understood in Lodash for instance) all intermediate function calls serve just for passing arguments while the only function that does some actual job after collecting all arguments is the last one.
I would call myFunc just "method of factory".
It's also a higher order function as #ASDFfgerte correctly points out in comment.
function functionOne(x){console.log(x);};
function functionTwo(var1) {
};
functionTwo(functionOne(2));
why does functionTwo work there ? it doesn't suppose work, does it?
because there is not an operation.
functionTwo(functionOne(2));
This means "immediately call functionOne, passing in 2. Then pass the result into functionTwo". So functionOne does its thing, logging out 2, and then returns undefined. And then undefined is passed into functionTwo.
Instead, if you're trying to experiment with callbacks you need to pass a function in, as in:
functionTwo(() => functionOne(2));
Once you do that, you'll no longer see a console.log unless you add some code to functionTwo.
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.
I'm going through Eloquent Javascript: Higher Order Functions example below and already read questions and answers here and here. But I'm still very confused.
function noisy(f) {
return function(arg) {
console.log("calling with", arg);
var val = f(arg);
console.log("called with", arg, "- got", val);
return val;
};
}
noisy(Boolean)(0);
// → calling with 0
// → called with 0 - got false
How can (0) be passed into noisy(f) since noisy() only takes one parameter and that is (Boolean)? I can see the inner function f(arg) is basically Boolean(0), but I don't understand how two parameters can get passed into a function that only allow one parameter. Would "noisy(Boolean)(0)(1)(2)(3);" be a valid function call? If so, how would you differentiate each value after Boolean within the noisy function? Which value will be referenced by "arg"?
The book noted the example function is modifying another function. Which function is being modified? I'm not understanding what the author meant by "modified".
but I don't understand how two parameters can get passed into a
function that only allow one parameter
noisy returns a function, Boolean is passed to noisy, 0 is passed to anonymous function returned from noisy, where f is Boolean, val becomes Boolean(0).
For example
function fn1(arg1) {
return function fn2(arg2) {
console.log(arg1, arg2)
}
}
// Call fn1, inside fn1 fn2 is called with `"b"` as parameter.
fn1("a")("b") // `a b`, `fn2`
This is the concept of currying in JavaScript where you can curry functions to return partially applied functions or pass in other functions
How can (0) be passed into noisy(f) since noisy() only takes one parameter and that is (Boolean)?
The answer to this is the curried function noisy() which expects a function f as parameter and returns another function. The returned function has a closure over noisy and as a result it can identify that Boolean was passed as parameter to noisy even after it was returned. That's why calling noisy(Boolean)(0) basically substitutes f=Boolean, arg=0
Refer this for more on currying: http://javascript.crockford.com/www_svendtofte_com/code/curried_javascript/ and closures: https://developer.mozilla.org/en/docs/Web/JavaScript/Closures
im learning javascript, and i have been following some video tutorial on youtube
this is the original code
function add(first, second, callback){
console.log(first+second);
callback();
}
function logDone(){
console.log("done");
}
add(2,3,logDone);
and the result of code on above is
5
main.js (line 4)
done
main.js (line 9)
and i make slight change to the code into this
function add(first, second, callback){
console.log(first+second);
callback;
}
function logDone(){
console.log("done");
}
add(2,3,logDone());
and the result is this
done
main.js (line 9)
5
main.js (line 4)
my question are:
could you explain to me why i got the result pile up that way?,
and what the difference if we call a function with the bracket () and without the bracket ()?
Explanation of the first snippet
function add(first, second, callback) {
console.log(first + second);
callback(); // run whatever the "callback" function is
}
function logDone() {
console.log("done");
}
add(2, 3, logDone); // pass in a function (not an invocation of a function) the
// function isn't run here
Explanation of the second snippet
function add(first, second, callback) {
console.log(first + second);
callback; // display the value of whatever "callback" is
}
function logDone() {
console.log("done");
}
add(2, 3, logDone()); // run "logDone" and then pass the result (which in this
// case is undefined) into add
As you can see, the first code snippet doesn't actually run the callback until in the add function, whereas the second snippet runs the callback before add so that it can pass whatever comes back from logDone into add.
Perhaps it becomes more clear to you when I alter the logDone declaration to this:
var logDone = function() {
console.log("done");
}
The identifier logDone basically is just a variable that references a function. To execute (also: call, or: invoke) the function you add parentheses: logDone().
So, in your first example you are merely passing the function itself as the third argument to add(), which then gets executed inside add(), with callback();.
In your second example, however, you are executing the function with logDone() immediately, which results in the return value of logDone()1 being passed as the third argument to the add() call. In other words, first logDone is executed (resulting in the first log message), then add is executed (resulting in the second log message).
Furthermore, the statement callback;, inside add, does not do anything. And if you would have left the parentheses just like in your first example, it would result in an error, because undefined2 is not a function.
1) which is undefined in this case, because logDone() does not explicitly return anything.
2) the value that was the result of the logDone() call.
and what the difference if we call a function with the bracket () and without the bracket ()?
If you have () then you call it. If you don't, then you don't.
could you explain to me why i got the result pile up that way?,
In the first example you pass a function to add and then the add function calls it.
In the second example, you call the function and pass its return value (undefined) to add which then mentions it in a statement but doesn't do anything with it.
When you do func([args=optional]), func is getting invocated or called.
and what the difference if we call a function with the bracket () and
without the bracket ()?
We call the function only when we do ()(arguments optional). When a function is without parenthesis, you're just using it's reference.
By storing a reference do a function in a variable, you can call it later, that's what callback is doing.
In the second snippet, since nothing is returned, callback will have undefined. Try calling it by doing callback() in the second snippet, and you should see an error stating,
undefined is not a function.
In the first example, you are passing a function as a parameter.
In your second example, you are passing a functions result after invocation as a parameter.