What is .map() doing in this situation?

Question

Using the Chrome Console, this is my input and output:

[0].map(Array);  [[0, 0, [0]]]; // output 

What is happening here?

EDIT

The reason this makes me curious is because something like

[0].map(String); 

Will return

["0"]; 

And not

["0", "0", "String"] 

Answer 1

The .map() function is calling the Array() function with three arguments, the value of the array element which is 0, the index of that element, also 0, and a reference to the whole array.

So it's like doing this:

var a = [0]; var index = 0 Array(a[index], index, a);   // create array with three elements 

The array returned by Array() then becomes the first element of the array that .map() creates, hence the extra level of nesting in your [[0, 0, [0]]] result.

EDIT regarding your edit: when you say [0].map(String); that results in String() being called with the same three arguments like String(a[index], index, a), but the String() function ignores all but the first argument, whereas Array() uses all supplied arguments.

Answer 2

Firstly, Array could be used as a function to create arrays:

var arr = Array(1, 2, "Hello");    console.log(arr); // [1, 2, "Hello"]

Secondly, map passes three parameters to its callback: the element, its index from the array and the array itself.

So, since your array contains one element, the line:

[0].map(Array); 

is equivalent to:

[Array(0, 0, [0])];     // the element 0 in the original array will be mapped into Array(0, 0, [0])