javascript - 删除具体是如何工作的？

Question

请检查jsfiddle - http://jsfiddle.net/du8svaym/

var a = [2, 4, "bang", undefined, NaN, 5];

for (i in a)
alert(a[i]); //alerting a[3]=undefined

delete a[1];
for (i in a)
alert(a[i]); //Why not alerting a[1]= undefined?

alert(a[1]); //it is undefined! but not alerted, what is happening under the hood?

如果您注意到，第一个循环警报会警报未定义的值。在第二个循环中，警报 a[1] 未定义，因为我们删除了它，但没有发出警报。两个 undefine 之间有什么区别，delete 设置 undefined 究竟有何不同？

Answer 1

delete 运算符从对象中删除属性。数组是 JavaScript 对象，如下所示:

var foo = {
  '0': 2,
  '1': 4,
  '2': "bang",
  '3': undefined, 
  '4': NaN,
  '5': 5
}
delete foo[1]; // completely removes the property '1' from foo.
console.log(foo);
// outputs:
{
  '0': 2,
  '2': "bang",
  '3': undefined, 
  '4': NaN,
  '5': 5
}