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Home/ Questions/Q 8756339
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T14:03:31+00:00 2026-06-13T14:03:31+00:00

I’m trying to learn Javascript by reading the source code of Fabric.js. In file

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I’m trying to learn Javascript by reading the source code of Fabric.js.

In file util/lang_array.js there is a method that looks like this:

    var slice = Array.prototype.slice;

    function invoke(array, method) {
        var args = slice.call(arguments, 2), result = [ ];

        for (var i = 0, len = array.length; i < len; i++) {
            result[i] = args.length ? array[i][method].apply(array[i], args) : array[i][method].call(array[i]);
        }
        return result;
    }

I don’t understand the statement var args = slice.call(arguments, 2). The particular part that puzzles me is passing 2 as the second argument (considering arguments being [array, method] then arguments[2] should be null). I looked up the JS reference and came to the conclusion that it basically initializes args to an empty array.

Then why not simply var args = []?

Thanks.

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  1. Editorial Team
    Editorial Team
    2026-06-13T14:03:33+00:00Added an answer on June 13, 2026 at 2:03 pm

    This line is there to pick up any extra arguments supplied to the function.

    Those extra arguments are then passed to the specified method in the .apply call.

    NB: JS does not check the formal parameters of functions. Omitted parameters will be undefined, and extra parameters will be (mostly) ignored, but available through the arguments pseudo-array.

    I’m unsure why the code bothers checking args.length to see whether to call .apply or .call – it looks like premature optimisation.

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