How to avoid the "use the array literal notation" jslint error in the following javascript code

Question

Possible Duplicate:
What's wrong with var x = new Array();

if (!Number.prototype.toZeroPaddedString) {
  Number.prototype.toZeroPaddedString = function (count) {
    "use strict";
    var str = this.toString();
    return (new Array(count + 1 - str.length)).join('0') + str;
  };
}

I want my code to be clean by jsLint standards, however, I just cannot imagine how to get rid of this error given what I want to do.

Any advice?

Thanks.

Not at all. How exactly do you plan to apply the answer to the question you mention in my case? — mark, Sep 02 '12 at 03:27
And make sure you tidy your script with tabs or 4 spaces instead 2 spaces — Elieder, Sep 02 '12 at 03:33
Elieder - that is a) personal preference and b) tabs suck on SO on iDevices and 4 spaces is a waste of space on any mobile/tablet so I strongly disagree! — mplungjan, Sep 02 '12 at 04:19

jeremy · Answer 1 · 2012-09-02T03:39:39.537

5

The array literal notation is an alternative to using new Array() because Crockford (rightfully) doesn't like the new keyword. The notation is [], just as the literal notation for new Object() is {}. To make it so that the length of the array is as desired, you can set .length on an array.

return (new Array(count + 1 - str.length)).join('0') + str;

To:

var arr = [];
arr.length = count + 1 - str.length;
return arr.join('0') + str;

edited Sep 02 '12 at 03:39

answered Sep 02 '12 at 03:22

jeremy

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I may be dead wrong, but I believe `new Array(someNumber)` is different from `new Array(elem1, elem2, ...)` in that it creates an array of *length* `someNumber` with undefined elements. – Sep 02 '12 at 03:27
`new Array(2) === [undefined, undefined]` whereas `[2]` is just `[2]`. Your answer seems to be simply misleading. – mark Sep 02 '12 at 03:28
Sorry for the confusion @delnan, I've updated my post, Mark. – jeremy Sep 02 '12 at 03:34
Now that's what I call misuse of the comma operator. Why not split that up in separate statements? Also, you're missing a `var`, your version makes `arr` global. – Sep 02 '12 at 03:38
@delnan I've updated again. Setting `var` wasn't working as I expected, so I set it as a global. – jeremy Sep 02 '12 at 03:40

score 0 · Accepted Answer · answered Sep 02 '12 at 03:38

Try this:

if (!Number.prototype.toZeroPaddedString) {
  Number.prototype.toZeroPaddedString = function (count) {
    "use strict";
    var str = this.toString();
    var ary = [];
    ary.length = count + 1 - str.length;
    return ary.join('0') + str;
  };
}

How to avoid the "use the array literal notation" jslint error in the following javascript code

2 Answers2