奇怪的 javascript 运算符:expr >>>0

Strange javascript operator: expr gt;gt;gt; 0(奇怪的 javascript 运算符:expr gt;gt;gt;0)

问题描述

以下函数旨在实现 IE 中的 indexOf 属性.如果你曾经不得不这样做，我相信你以前见过.

the following function is designed to implement the indexOf property in IE. If you've ever had to do this, I'm sure you've seen it before.

if (!Array.prototype.indexOf){

  Array.prototype.indexOf = function(elt, from){

    var len = this.length >>> 0;
    var from = Number(arguments[1]) || 0;

    from = (from < 0)
         ? Math.ceil(from)
         : Math.floor(from);

    if (from < 0)
      from += len;

    for (; from < len; from++){
      if (from in this &&    
          this[from] === elt)
        return from;
    }

    return -1;    
  };
}

我想知道是否像作者在初始长度检查中所做的那样使用三个大于号?

I'm wondering if it's common to use three greater than signs as the author has done in the initial length check?

var len = this.length >>>0

在控制台中执行此操作只会返回我传递给它的对象的长度，而不是 true 或 false，这让我思考语法的目的.这是一些我不知道的高级 JavaScript Ninja 技术吗?如果有，请赐教！

Doing this in a console simply returns the length of the object I pass to it, not true or false, which left me pondering the purpose of the syntax. Is this some high-level JavaScript Ninja technique that I don't know about? If so, please enlighten me!

推荐答案

来源:链接

这是零填充右移移位二进制的运算符第一个操作数的表示名额的权利由第二个操作数指定.位向右移动被丢弃和零被添加到左边.用正数你会得到结果与符号传播右移运算符，但负数失去符号像接下来一样变得积极例如，其中(假设 'a' 是-13) 将返回 1073741820:

This is the zero-fill right shift operator which shifts the binary representation of the first operand to the right by the number of places specified by the second operand. Bits shifted off to the right are discarded and zeroes are added on to the left. With a positive number you would get the same result as with the sign-propagating right shift operator, but negative numbers lose their sign becoming positive as in the next example, which (assuming 'a' to be -13) would return 1073741820:

代码:

result = a >>> b;

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