javascript array.sort带有未定义的值 [英] javascript array.sort with undefined values

问题描述

Array.prototype.sort 如何处理数组中未定义的值？

  var array = [1，undefined，2，undefined，3，undefined，4]; 
 var array2 = []; 
 array2 [0] = 1; array2 [2] = 2; array2 [4] = 3; array2 [6] = 4; 
  

当调用 array.sort（function（l，r）{.. 。}）; 值 undefined 从不传递为 l 或 r 。

我可以保证所有未定义的值会一直到所有浏览器的数组末尾？

以下循环处理数组中所有非 undefined 数据

  array.sort（）; 
 for（var i = 0; array [i]！== undefined; i ++）{
 // handle array 
} 
  

您可以假设没有人声明未定义为变量。

是的，你可以放心地假设 undefined 将被移动到数组的末尾。

从 MDC ：

在JavaScript 1.2中，此方法不再将未定义的元素转换为null;而是将它们排序到数组的高端

从 spec，15.4.4.11
：

因为不存在的属性值总是比大于未定义的属性值，而
undefined总是比任何其他值大，未定义的属性值总是排序到结果的结尾，
后跟不存在的属性值

How does Array.prototype.sort handle undefined values in an array?

var array = [1,undefined,2,undefined,3,undefined,4];
var array2 = [];
array2[0] = 1;array2[2] = 2;array2[4] = 3;array2[6] = 4;

When calling array.sort(function(l,r) { ... }); The values undefined are never passed in as l or r.

Can I guarantee that all the undefined values will always go to the end of the array for all browsers?

Would the following loop handle all the non undefined data in an array

array.sort();
for (var i = 0; array[i] !== undefined; i++) {
    // handle array
}

You may assume that no-one declared undefined as a variable.

解决方案

Yes, you can safely assume undefined will get moved to the end of the array.

From MDC:

In JavaScript 1.2, this method no longer converts undefined elements to null; instead it sorts them to the high end of the array

From the spec, 15.4.4.11 :

Because non-existent property values always compare greater than undefined property values, and undefined always compares greater than any other value, undefined property values always sort to the end of the result, followed by non-existent property values.

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