in this higher level function, where does the argument, n come from?

Question

In the following code from ch 5 of Eloquent Javascript , where does that value of the argument n come from?

Why does this function return anything . I guess, I hate to ask a unspecific, cliche question, but I am baffled: How does this function work?

 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 

Answer 1

As you can see n is defined as a parameter in the callback here:

repeat(3, function(n) {
//                 ^

Let's go into the repeat function:

function repeat(times, body) {
    for (var i = 0; i < times; i++) body(i);
    // We're calling our callback with i ^
}

Here times will be 3 and body will be our anonymous function (callback). Hence when we call body(i) we actually call the following (switched n out with i , as we're calling body(i) ):

unless(i % 2, function() {
    console.log(i, "is even");
})

---

Here's your entire source unwrapped:

var times = 3;

for (var i = 0; i < times; i++) {
    var n = i; // We're renaming "i" (from body(i)) to "n" (in function(n))

    if (!(n % 2)) {
        console.log(n, "is even");
    }
}

Answer 2

The variable n is defined in the callback method you are passing to "repeat":

repeat(3,function(n){

unless otherwise modified inside that callback function (or defined using "var n = ..." that value will persist in the scope all the way to the console.log call.