What is the difference between “new Number(…)” and

问题:

In Javascript, one of the reliable ways to convert a string to a number is the Number constructor:

var x = Number('09'); // 9, because it defaults to decimal

Inspired by this question, I started wondering — what is the difference between the above and:

var x =new Number('09');

Number certainly looks better, but it seems like a slightly inappropriate use of a constructor. Are there any side effects or any difference to using it without the new? If there is no difference, why not, and what is the purpose of new?

回答1:

In the first case, you are using the Number Constructor Called as a Function, as described in the Specification, that will simply perform a type conversion, returning you a Number primitive.

In the second case, you are using the Number Constructor to make a Number object:

var x = Number('09');
typeof x; // 'number'

var x = new Number('09');
typeof x; // 'object'

Number('1') === new Number('1'); // false

The difference may be subtle, but I think it's important to notice how wrapper objects act on primitive values.

回答2:

Number returns a primitive number value. Yeah, it's a bit odd that you can use a constructor function as a plain function too, but that's just how JavaScript is defined. Most of the language built-in types have weird and inconsistent extra features like this thrown in.

new Number constructs an explicit boxed Number object. The difference:

typeof Number(1)      // number
typeof new Number(1)  // object

In contrast to Java's boxed primitive classes, in JavaScript explicit Number objects are of absolutely no use.

I wouldn't bother with either use of Number. If you want to be explicit, use parseFloat('09'); if you want to be terse, use +'09'; if you want to allow only integers, use parseInt('09', 10).

回答3:

SpiderMonkey-1.7.0:

js> typeof new Number('09');
object
js> typeof Number('09');
number

回答4:

Number (without new) doesn't seem to result exactly in a primitive. In the following example the anyMethod() is called (if in the Number prototype).

Number(3).anyMethod()

Whereas

3.anyMethod()

will not work.