understanding callback and return functions in javascript

Question

I am new to programming and trying to understand callback functions and functions in general. This program compares 2 values passed in the functions(using callback) and return true/false to us.

function func1(param1, callback) {
    return callback(param1);
}

function func2(param2) {
    return function(param3) {
        return param3 > param2;
     }
}

var functionResult = func1(10, func2(9)); 
console.log(functionResult);       // prints - true

Question - In this program above, how does the return function inside the func2 function, return the value directly to us, without being invoked? I thought in this line var functionResult = func1(10, func2(9)); func2(9) will return only the text

function(param3) {
        return param3 > param2;
     }

and then I would have to invoke it again with ().

Possible duplicate of How do JavaScript closures work?

Aaron Hudon
– Aaron Hudon

2018-11-14 22:23:42 +00:00
Commented Nov 14, 2018 at 22:23 — Aaron Hudon
– Aaron Hudon, Commented Nov 14, 2018 at 22:23

Felix Kling · Accepted Answer · 2018-11-14 19:40:48Z

2

how does the return function inside the func2 function, return the value directly to us, without being invoked?

It is invoked. func1 invokes it here:

callback(param1)

func2(9) will return only the text ...

That's not text, that's actually a function (object). It is passed to func1 which in turn calls it and returns its return value.

and then I would have to invoke it again with ().

Yes, which, again, is what func1 does:

callback(param1)

Lets take things apart:

 var functionResult = func1(10, func2(9));

is the same as

var func2Result = func2(9);
var functionResult = func1(10, func2Result);

Since we know what func1 does, we can replace the call to it with it's implementation. 10 is passed as param1 and func2Result is passed as callback, so the code becomes:

var func2Result = func2(9);
var functionResult = func2Result(10); // callback(param1);

Now you should be able to see that the return value of func2 is actually called.

edited Nov 14, 2018 at 19:40

answered Nov 14, 2018 at 19:35

Felix Kling

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3 Comments

roan shaz Over a year ago

thanks, the taking things apart part really made it clear to me

roan shaz Over a year ago

I just realized, the code is essentially func2(9)()

Felix Kling Over a year ago

Yes, you can also omit func2Result... maybe I should have done that instead, but I thought maybe seeing ()() is also confusing ¯\_(ツ)_/¯

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