Regarding with JavaScript array prototype

Question

I have recently read tutorial associated with JavaScript from this page, I tried some piece of code but i didn't understand some logic. Below is author code.

function itself:

var lz="";
lz.memo = function (fn) {
var cache = {};
return function () {
    var key = [].join.call(arguments, '§') + '§';
    if (key in cache) {
        return cache[key];        }
    return cache[key] = fn.apply(this, arguments);
  };
};

and its execution

 var foo = 1;
function bar(baz) {
    return baz + foo;
}
var cached = lz.memo(bar);
console.log(cached(1));
foo += 1;
console.log(cached(1)); //2

But if I change

var key = [].join.call(arguments, '§') + '§';

to

var key=arguments[0];

it also works (Caching works). What is the purpose here the author used

var key = [].join.call(arguments, '§') + '§';

Thank for attention! Here is CODEPEN of code

T.J. Crowder · Accepted Answer · 2020-06-29T08:20:31.880

it also works (Caching works)

Only because you called it with just one argument. (And it didn't quite work the same way, the cache key will be missing the § on the key that the original code had.)

What is the purpose here the author used

They're creating a string with all of the arguments you supply joined together with § between them and § at the end, like this:

function foo() {
  return [].join.call(arguments, '§') + '§';
}
console.log(foo(1)); // 1§
console.log(foo(1, 2)); // 1§2§
console.log(foo("a", "b", "c")); // a§b§c§

They're doing it because arguments is array-like, but it isn't an array, and so it doesn't have the join method from Array.prototype.

There's actually no reason for the temporary array they create, you can just call join directly:

return Array.prototype.join.call(arguments, '§') + '§';

Another opton is to convert arguments to an actual array:

return Array.prototype.slice.call(arguments).join('§') + '§';

Or in ES2015 (or with a shim):

return Array.from(arguments).join('§') + '§';

here foo variable is defined as global. If i define foo next to key variable, function throws undefined foo variable. How can initialize and assign foo variable next key so that function can add? — Azer Qurbanov, Mar 07 '16 at 10:22
@AzerQurbanov: Sorry, I couldn't understand that. To the extent I could, though, it sounds like a new question. If so, post a new question. :-) — T.J. Crowder, Mar 07 '16 at 10:32

Regarding with JavaScript array prototype

1 Answers1