javascript - Why should you not use Number as a constructor?

Question

I entered this statement in JSLint:

var number = new Number(3);

And received the following message:

Do not use Number as a constructor.

Why is that? The statement is creating a number object, not a primitive value, so I don't see why using new is a problem.

EDIT: Thanks for all the responses. They've got me thinking further, so I posted a follow-up question here.

score 39 · Accepted Answer

除了打破 === 和 typeof 返回“object”之外，使用 Number 构造函数还改变了值在布尔上下文中的使用方式。由于“new Number(0)”是一个对象，而不是文字值，它的计算结果为“true”，因为它不为空。例如：

var n1 = 0;
var n2 = new Number(0);

n1 == n2  // true
n1 === n2 // false
if (n1) {
    // Doesn't execute
}
if (n2) {
    // Does execute, because n2 is an object that is not null
}

比在数字文字和数字对象之间打破 === 更糟糕的是， == 甚至在两个数字对象之间都不起作用（至少不是以直观的方式——它们测试的是同一性，而不是相等性）。

var n1 = new Number(3);
var n2 = new Number(3);

alert(n1 == n2); // false
alert(n1 === n2); // false

score 19 · Accepted Answer

var number = new Number(3);
alert(typeof number); // gives "object"

Making the variable number have a type of Object is probably not the most desired outcome. Whereas:

var number = Number(3);
alert(typeof number); // gives "number"

score 6 · Accepted Answer

Unfortunately, the JSLint docs don't go into any further detail than "does not expect to see", so we're left to guess. My own suspicion is that this is to make type-checking easier:

assert(typeof 3             === "number");
assert(typeof new Number(3) === "object");

If you mix the two in your code, your type checks become more complex:

if (typeof foo === "number" || foo instanceof Number) { … }

However, JSLint also takes issue with the Object and Array constructors, which do not make this distinction, so it may simply be the author's coding-style preference:

assert(typeof []           === "object");
assert(typeof new Array()  === "object");
assert(typeof {}           === "object");
assert(typeof new Object() === "object");

Edit: Steven's answer raises an excellent point — the non-typecasting equality operator (===). Number objects and number primitives will never be considered equal by this operator, even if their values are the same:

assert(3 !== new Number(3));

score 6 · Accepted Answer

new Number() does not return the same object as a number literal. This means that using new Number() breaks ===, which is the best way to check for exact equality in Javascript.

>>> 3 == 1 + 2
true
>>> 3 === 1 + 2
true
>>> new Number(3) == 1 + 2
true
>>> new Number(3) === 1 + 2
false

You can find the rationale for JSLint's behavior in the author's book, JavaScript: The Good Parts, in Appendix C.

score 5 · Accepted Answer

它速度较慢，并且需要更多内存。运行时可以将不可变文字视为不可变文字。这意味着当它遇到3代码中的某个地方时，它可以将其优化为共享对象。使用Number构造函数时，会为每个实例分配新内存。

score 1 · Accepted Answer

在 JavaScript 中，一个 Object 类型不等于另一个 Object 类型，即使它们具有完全相同的值，除非它们都是 EXACT SAME 对象。

换句话说，在下面 Matthew 的示例中，n1 == n2 为假，因为您将两个 REFERENCES 与两个 SEPARATE 对象进行比较，但 n1 == n1 为真，因为您正在比较对 EXACT SAME 对象的引用。

所以，虽然我现在明白了为什么使用 Number 作为构造函数会导致问题，但我发现在比较 Number 对象时可以使用 valueOf 属性。

换句话说，n1.valueOf == n2.valueOf 是真的！（这是因为您正在比较 valueOf FUNCTION 的返回值，而不是对象本身的 REFERENCES。）

_{这个答案/摘要是从它不属于的问题中提取的。}

javascript - Why should you not use Number as a constructor?

6 回答 6

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