CODEWITHSUNDEEP
javascripthigher-order-functions
Alexander
Alexander

Reputation: 255

Parameters handling in JavaScript functions

Currently I'm reading JavaScript book. There is a code snippet in it which I can't understand. What's happening in the line repeat(3, function(n) {? Why we can pass parameter n to the second argument of the function repeat, because in its declaration there is nothing about passing parameters? How does repeat understand that it should pass parameter n to the unless function?

function unless(test, then) {
    if (!test) then();
}
function repeat(times, body) {
    for (var i = 0; i < times; i++) body(i);
}
repeat(3, function(n) {
    unless(n % 2, function() {
        console.log(n, "is even");
    });
});
// → 0 is even
// → 2 is even

Upvotes: 2

Views: 47

Answers (3)

nem035
nem035

Reputation: 35491

Why we can pass parameter n to the second argument of the function repeat, because in its declaration there is nothing about passing parameters

You are not passing n as the second argument to repeat(), you are passing an anonymous function that takes a single parameter and you chose to name its parameter n (so parameter of the function that is passed in)

Functions in JavaScript are, in simple words, just objects that can also be executed. This means you can pass functions around as parameters to other functions, or add properties to them like you would to objects, etc.

Here's an illustration of what is happening in your example:

Function repeat is defined with two arguments:

repeat(times, body)

So, all you are doing is passing a function as the body argument. Writing it like this is equivalent:

var times = 3;
var body = function(n) {
    unless(n % 2, function() {
        console.log(n, "is even");
    });
};

repeat(time, body);

How does repeat understand that it should pass parameter n to the unless function?

As you can see above, repeat is not passing anything to unless(). It is your anonymous function (stored in body in my example above) that is actually passing n to unless.

Upvotes: 2

jaggedsoft
jaggedsoft

Reputation: 4038

If you simply want to write a function to repeat x number of times, and check if it's even, you'd probably want to do it like this:

function repeat(times) {
    for (var i = 0; i < times; i++) {
        output.innerHTML+= is_even(i)+"\n";
    }
}
function is_even(n) {
   if ( n % 2 ) return n+" is even";
   return n+" is odd";
}
var output = document.getElementById('output');
repeat(6); // Output variable wasn't passed to this function, we're using it globally
<pre id='output'></pre>

Upvotes: 0

Justin Niessner
Justin Niessner

Reputation: 245429

You're not passing a parameter n at all.

In reality, you're passing an entire anonymous function as a parameter (functions are first-class citizens in JavaScript and can be passed around just like other variables).

If you look, the function is passed as the body parameter to the method repeat. repeat then calls the function body with the parameter of i...which is the parameter n in the anonymous function.

Upvotes: 2

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