JavaScript puzzle to reimplement Array.prototype.slice not working as expected

Question

This is not for a job interview, don't worry :)

I am curious as to why these two functions return differently:

var redundantSlice1 = function() {
    return arguments[0].slice(Array.prototype.slice.call(arguments).slice(1).join())
};

var redundantSlice2 = function() {
    return arguments[0].slice(1,2)
};

redundantSlice1([1,2,3], 1, 2) // [1,2,3]

redundantSlice2([1,2,3], 1, 2) // [2]

redundantSlice2 returns what I expect, but the 1,2 are hardcoded, this is what I am trying to avoid.

Why does the first function have what I expect to be 1,2 as undefined?

.join returns a string. So the first example is equivalent to arr.slice("1,2"). That's very different from arr.slice(1, 2). I'm not sure how .slice treats invalid arguments (apparently it treats it like 0 or nothing was passed). — Felix Kling
– Felix Kling, Commented Nov 8, 2015 at 2:06
Yeah, that's what I found. I had the same results passing null and undefined. — Jazzy
– Jazzy, Commented Nov 8, 2015 at 2:07
you need to look at function.apply - but as slice only ever takes 0, 1 or 2 arguments, I'd define your wrapper function as function(array, begin, end) then simply return [].slice.call(array, begin, end) - that's just as much a re-implementation as you've written — Jaromanda X
– Jaromanda X, Commented Nov 8, 2015 at 2:13
There are more constraints to the problem that I didn't want to give away as to not cheat, but @FelixKling answered my initial question. It's a string. — Jazzy
– Jazzy, Commented Nov 8, 2015 at 2:26

Austin Brunkhorst · Accepted Answer · 2015-11-08 02:19:54Z

1

In redundantSlice1, .join( ) converts the arguments into a string with , as the delimiter by default. So with your sample, the statement

Array.prototype.slice.call( arguments ).slice( 1 ).join( )

Is actually "1,2" not [ 1, 2 ] like you're expecting.

In addition to this, you're only passing one argument to .slice( ). In order to match the behavior of redundantSlice2, you must use .apply( ).

var redundantSlice1 = function() {
    var firstArg = arguments[ 0 ];

    return Array.prototype.slice.apply(
        firstArg,
        Array.prototype.slice.call( arguments, 1 )
    );
};

edited Nov 8, 2015 at 2:19

answered Nov 8, 2015 at 2:13

Austin Brunkhorst

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3 Comments

Felix Kling Over a year ago

This could probably also just be done with arguments[0].slice(arguments[1], arguments[2]).

Jaromanda X Over a year ago

what if arguments[0] is "array-like"?

Austin Brunkhorst Over a year ago

@FelixKling Yep, I'm just being consistent with the OP's approach.

Jaromanda X · Accepted Answer · 2015-11-08 02:18:08Z

1

If you want to do this so you can pass an array or an array-like object and get the same result

var redundantSlice1 = function(array, begin, end) {
    return Array.prototype.slice.call(array, begin, end)
};

this will work with 1, 2 or 3 arguments passed in exactly like Array.slice works

If you insist on slicing the arguments and have no "formal parameters" to the function

var redundantSlice1 = function() {
    return Array.prototype.slice.apply(arguments[0], Array.prototype.slice.call(arguments, 1));
};

answered Nov 8, 2015 at 2:18

Jaromanda X

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