奇怪的javascript运算符：expr>>> 0 [英] Strange javascript operator: expr >>> 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，我相信你以前见过它。

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;    
  };
}

我想知道作为作者使用三个大于标志是否常见在初始长度检查中做了什么？

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，这让我想到了语法的目的。这是一些我不了解的高级JavaScript忍者技术吗？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： LINK

这是零填充右移
运算符，它将第一个操作数的二进制
表示转换为
右边第二个操作数指定的位数
。转移到右边的
比特被丢弃
并在左边添加零。
如果是正数，你将获得
与
符号传播右移运算符相同的结果，
但是负数会丢失他们的符号
变为正数下一个
的例子，（假设'a'为
-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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