JavaScript "new Array(n)" and "Array.prototype.map" weirdness

JavascriptArraysMap Function

Javascript Problem Overview


I've observed this in Firefox-3.5.7/Firebug-1.5.3 and Firefox-3.6.16/Firebug-1.6.2

When I fire up Firebug:

var x = new Array(3)
console.log(x) 
// [undefined, undefined, undefined]

var y = [undefined, undefined, undefined]
console.log(y) 
// [undefined, undefined, undefined]

console.log( x.constructor == y.constructor) // true

console.log( 
  x.map(function() { return 0; })
)
// [undefined, undefined, undefined]

console.log(
  y.map(function() { return 0; })
)
// [0, 0, 0]

What's going on here? Is this a bug, or am I misunderstanding how to use new Array(3)?

Javascript Solutions


Solution 1 - Javascript

I had a task that I only knew the length of the array and needed to transform the items. I wanted to do something like this:

let arr = new Array(10).map((val,idx) => idx);

To quickly create an array like this:

[0,1,2,3,4,5,6,7,8,9]

But it didn't work because: (see Jonathan Lonowski's answer a few answers beneath)

The solution could be to fill up the array items with any value (even with undefined) using Array.prototype.fill()

let arr = new Array(10).fill(undefined).map((val,idx) => idx);

console.log(new Array(10).fill(undefined).map((val, idx) => idx));

Update

Another solution could be:

let arr = Array.apply(null, Array(10)).map((val, idx) => idx);

console.log(Array.apply(null, Array(10)).map((val, idx) => idx));

Solution 2 - Javascript

It appears that the first example

x = new Array(3);

Creates an array with a length of 3 but without any elements, so the indices [0], [1] and [2] is not created.

And the second creates an array with the 3 undefined objects, in this case the indices/properties them self are created but the objects they refer to are undefined.

y = [undefined, undefined, undefined]
// The following is not equivalent to the above, it's the same as new Array(3)
y = [,,,];

As map runs on the list of indices/properties, not on the set length, so if no indices/properties is created, it will not run.

Solution 3 - Javascript

With ES6, you can do [...Array(10)].map((a, b) => a) , quick and easy!

Solution 4 - Javascript

ES6 solution:

[...Array(10)]

Doesn't work on typescript (2.3), though

Solution 5 - Javascript

From the MDC page for map:

> [...] callback is invoked only for indexes of the array which have assigned value; [...]

[undefined] actually applies the setter on the index(es) so that map will iterate, whereas new Array(1) just initializes the index(es) with a default value of undefined so map skips it.

I believe this is the same for all iteration methods.

Solution 6 - Javascript

The arrays are different. The difference is that new Array(3) creates an array with a length of three but no properties, while [undefined, undefined, undefined] creates an array with a length of three and three properties called "0", "1" and "2", each with a value of undefined. You can see the difference using the in operator:

"0" in new Array(3); // false
"0" in [undefined, undefined, undefined]; // true

This stems from the slightly confusing fact that if you try to get the value of a non-existent property of any native object in JavaScript, it returns undefined (rather than throwing an error, as happens when you try to refer to a non-existent variable), which is the same as what you get if the property has previously been explictly set to undefined.

Solution 7 - Javascript

For reasons thoroughly explained in other answers, Array(n).map doesn't work. However, in ES2015 Array.from accepts a map function:

let array1 = Array.from(Array(5), (_, i) => i + 1)
console.log('array1', JSON.stringify(array1)) // 1,2,3,4,5

let array2 = Array.from({length: 5}, (_, i) => (i + 1) * 2)
console.log('array2', JSON.stringify(array2)) // 2,4,6,8,10

Solution 8 - Javascript

In ECMAScript 6th edition specification.

new Array(3) only define property length and do not define index properties like {length: 3}. see https://www.ecma-international.org/ecma-262/6.0/index.html#sec-array-len Step 9.

[undefined, undefined, undefined] will define index properties and length property like {0: undefined, 1: undefined, 2: undefined, length: 3}. see https://www.ecma-international.org/ecma-262/6.0/index.html#sec-runtime-semantics-arrayaccumulation ElementList Step 5.

methods map, every, some, forEach, slice, reduce, reduceRight, filter of Array will check the index property by HasProperty internal method, so new Array(3).map(v => 1) will not invoke the callback.

for more detail, see https://www.ecma-international.org/ecma-262/6.0/index.html#sec-array.prototype.map

How to fix?

let a = new Array(3);
a.join('.').split('.').map(v => 1);

let a = new Array(3);
a.fill(1);

let a = new Array(3);
a.fill(undefined).map(v => 1);

let a = new Array(3);
[...a].map(v => 1);

Solution 9 - Javascript

I think the best way to explain this is to look at the way that Chrome handles it.

>>> x = new Array(3)
[]
>>> x.length
3

So what is actually happening is that new Array() is returning an empty array that has a length of 3, but no values. Therefore, when you run x.map on a technically empty array, there is nothing to be set.

Firefox just 'fills in' those empty slots with undefined even though it has no values.

I don't think this is explicitly a bug, just a poor way of representing what is going on. I suppose Chrome's is "more correct" because it shows that there isn't actually anything in the array.

Solution 10 - Javascript

Not a bug. That's how the Array constructor is defined to work.

From MDC:

> When you specify a single numeric parameter with the Array constructor, you specify the initial length of the array. The following code creates an array of five elements:

var billingMethod = new Array(5);

> The behavior of the Array constructor depends on whether the single parameter is a number.

The .map() method only includes in the iteration elements of the array that have explicitly had values assigned. Even an explicit assignment of undefined will cause a value to be considered eligible for inclusion in the iteration. That seems odd, but it's essentially the difference between an explicit undefined property on an object and a missing property:

var x = { }, y = { z: undefined };
if (x.z === y.z) // true

The object x does not have a property called "z", and the object y does. However, in both cases it appears that the "value" of the property is undefined. In an array, the situation is similar: the value of length does implicitly perform a value assignment to all the elements from zero through length - 1. The .map() function therefore won't do anything (won't call the callback) when called on an array newly constructed with the Array constructor and a numeric argument.

Solution 11 - Javascript

Just ran into this. It sure would be convenient to be able to use Array(n).map.

Array(3) yields roughly {length: 3}

[undefined, undefined, undefined] creates the numbered properties:
{0: undefined, 1: undefined, 2: undefined, length: 3}.

The map() implementation only acts on defined properties.

Solution 12 - Javascript

If you are doing this in order to easily fill up an array with values, can't use fill for browser support reasons and really don't want to do a for-loop, you can also do x = new Array(3).join(".").split(".").map(... which will give you an array of empty strings.

Quite ugly I have to say, but at least the problem and intention are quite clearly communicated.

Solution 13 - Javascript

Since the question is why, this has to do with how JS was designed.

There are 2 main reasons I can think of to explain this behavior:

  • Performance: Given x = 10000 and new Array(x) it is wise for the constructor to avoid looping from 0 to 10000 to fill the array with undefined values.

  • Implicitly "undefined": Give a = [undefined, undefined] and b = new Array(2), a[1] and b[1] will both return undefined, but a[8] and b[8] will also return undefined even if they're out of range.

Ultimately, the notation empty x 3 is a shortcut to avoid setting and displaying a long list of undefined values that are undefined anyway because they are not declared explicitly.

Note: Given array a = [0] and a[9] = 9, console.log(a) will return (10) [0, empty x 8, 9], filling the gap automatically by returning the difference between the two values declared explicitly.

Solution 14 - Javascript

Here's a simple utility method as a workaround:

Simple mapFor

function mapFor(toExclusive, callback) {
    callback = callback || function(){};
    var arr = [];
    for (var i = 0; i < toExclusive; i++) {
        arr.push(callback(i));
    }
    return arr;
};

var arr = mapFor(3, function(i){ return i; });
console.log(arr); // [0, 1, 2]
arr = mapFor(3);
console.log(arr); // [undefined, undefined, undefined]

Complete Example

Here's a more complete example (with sanity checks) which also allows specifying an optional starting index:

function mapFor() { var from, toExclusive, callback; if (arguments.length == 3) { from = arguments[0]; toExclusive = arguments[1]; callback = arguments[2]; } else if (arguments.length == 2) { if (typeof arguments[1] === 'function') { from = 0; toExclusive = arguments[0]; callback = arguments[1]; } else { from = arguments[0]; toExclusive = arguments[1]; } } else if (arguments.length == 1) { from = 0; toExclusive = arguments[0]; }

callback = callback || function () {};

var arr = [];
for (; from < toExclusive; from++) {
    arr.push(callback(from));
}
return arr;

}

var arr = mapFor(1, 3, function (i) { return i; }); console.log(arr); // [1, 2] arr = mapFor(1, 3); console.log(arr); // [undefined, undefined] arr = mapFor(3); console.log(arr); // [undefined, undefined, undefined]

Counting Down

Manipulating the index passed to the callback allows counting backwards:

var count = 3;
var arr = arrayUtil.mapFor(count, function (i) {
    return count - 1 - i;
});
// arr = [2, 1, 0]

Attributions

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Solution 1 - JavascriptcstuncsikView Answer on Stackoverflow
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Solution 3 - JavascriptManuel BeaudruView Answer on Stackoverflow
Solution 4 - JavascriptSerge InternView Answer on Stackoverflow
Solution 5 - JavascriptJonathan LonowskiView Answer on Stackoverflow
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