什么是x>>> 0呢？ [英] What does x >>> 0 do?

问题描述

可能重复：

零填充位移0有什么用呢？ （a>>> 0）

我一直在尝试我的一个项目中的一些函数式编程概念，我正在阅读 Array.prototype.map ，这是ES5中的新功能，如下所示：

I've been trying out some functional programming concepts in a project of mine and I was reading about Array.prototype.map, which is new in ES5 and looks like this:

Array.prototype.map = function(fun) {
    "use strict";
    if (this === void 0 || this === null) {
        throw new TypeError();
    }
    var t = Object(this);
    var len = t.length >>> 0;
    if (typeof fun !== "function") {
        throw new TypeError();
    }
    var res = new Array(len);
    var thisp = arguments[1];
    for (var i = 0; i < len; i++) {
        if (i in t) {
            res[i] = fun.call(thisp, t[i], i, t);
        }
    }
    return res;
};

我想知道为什么它正在做 t.length>> ;> 0 。因为它似乎没有做任何事情。 x>>> 0 // - > X ！ （只要x是一个数字，显然）

What I'm wondering is why it's doing t.length >>> 0. Because it doesn't seem to do anything. x >>> 0 //-> x! (as long as x is a number, obviously)

另外，请注意我不知道按位运算符是如何工作的。

Also, note that I don't know how bitwise operators work.

推荐答案

x>>> 0 执行0位的逻辑（无符号）右移，相当于无操作。但是，在右移之前，它必须将 x 转换为无符号的32位整数。因此， x>>>的整体效果0 将 x 转换为32位无符号整数。

x >>> 0 performs a logical (unsigned) right-shift of 0 bits, which is equivalent to a no-op. However, before the right shift, it must convert the x to an unsigned 32-bit integer. Therefore, the overall effect of x >>> 0 is convert x into a 32-bit unsigned integer.

这可确保 len 是非负数。

js> 9 >>> 0
9
js> "9" >>> 0
9
js> "95hi" >>> 0
0
js> 3.6 >>> 0
3
js> true >>> 0
1
js> (-4) >>> 0
4294967292

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