using a unique first argument in Javascript when calling the apply method

In this particular scenario, the output varies depending on the first argument passed to apply

The snippet of code looks like this:

var fruitarray = [];
fruitarray[0] = ['strawberry', 'orange'];
fruitarray[1] = ['lime', 'peach', 'banana'];
fruitarray[2] = ['tangerine', 'apricot'];

var array1= fruitarray.concat.apply([], fruitarray); // array1 becomes 'tangerine'

var array2= fruitarray.concat.apply('undefined', fruitarray); // array2 becomes 'banana'

var array3= fruitarray.concat.apply(fruitarray , fruitarray); // array3 becomes 'lime'

var array4= fruitarray.concat.apply(this, fruitarray); // array4 becomes 'banana'

What could be the reason behind this?

**** EDIT: *****

Apologies!!!

Here is the corrected code:

var fruitarray = [];
fruitarray[0] = ['strawberry', 'orange'];
fruitarray[1] = ['lime', 'peach', 'banana'];
fruitarray[2] = ['tangerine', 'apricot'];

var array1= fruitarray.concat.apply([], fruitarray); 
console.log(array1[5]); // 'tangerine'

var array2= fruitarray.concat.apply('undefined', fruitarray);
console.log(array2[5]); // 'banana'

var array3= fruitarray.concat.apply(fruitarray , fruitarray);
console.log(array3[5]); // 'lime'

var array4= fruitarray.concat.apply(this, fruitarray);
console.log(array4[5]); // 'banana'

I am currently using Firebug 2.0.9 for testing (Firefox)

Answer №1

Various arguments lead to distinct results!

The this parameter in the array concat function is significant as it indicates the array being concatenated with. Regardless of obtaining the function through

<em>fruitarray.concat</em>.apply(…
, you could have also opted for Array.prototype.concat.apply(… or [].concat.apply(….

So how can you achieve those outcomes now?

fruitarray.concat.apply([], fruitarray);
// equals
[].concat(['strawberry', 'orange'], ['lime', 'peach', 'banana'], ['tangerine', 'apricot']);
// resulting in array1
['strawberry', 'orange', 'lime', 'peach', 'banana', 'tangerine', 'apricot']

fruitarray.concat.apply(fruitarray, fruitarray);
// equals
fruitarray.concat(['strawberry', 'orange'], ['lime', 'peach', 'banana'], ['tangerine', 'apricot']);
// resulting in array3
[[…], […], […], 'strawberry', 'orange', 'lime', 'peach', 'banana', 'tangerine', 'apricot']

Now, what happens if a non-array item is passed (like a string or a Window object)? It simply gets encapsulated in a single-element array that is concatenated instead of being spread out.

fruitarray.concat.apply('undefined', fruitarray);
// equals
['undefined'].concat(['strawberry', 'orange'], ['lime', 'peach', 'banana'], ['tangerine', 'apricot']);
// resulting in array2
['undefined', 'strawberry', 'orange', 'lime', 'peach', 'banana', 'tangerine', 'apricot']

fruitarray.concat.apply(this, fruitarray);
// equals
[this].concat(['strawberry', 'orange'], ['lime', 'peach', 'banana'], ['tangerine', 'apricot']);
// resulting in array4
[{window}, 'strawberry', 'orange', 'lime', 'peach', 'banana', 'tangerine', 'apricot']

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