javascript - 删除未定义的数组元素和删除有什么区别？

Question

MDN 定义了以下差异，但我不理解差异。我 创建了一个数组，然后删除了一个元素，然后尝试调用该元素，但它返回了 undefined。

var ary=[0,1,2,3,4];
delete ary[3];
ary[3]// returns undefined
ary.length //returns 5

var ary2=[0,1,2,3,4];
ary2[2]=undefined
ary2[2]// returns undefined
ary2.length// returns 5

删除数组元素时，数组长度不受影响。即使您删除了数组的最后一个元素，这仍然成立。

当删除运算符删除数组元素时，该元素不再存在于数组中。在以下示例中，trees[3] 已通过 delete 删除。

var trees = ["redwood","bay","cedar","oak","maple"];
delete trees[3];
if (3 in trees) {
// this does not get executed
}

如果您希望数组元素存在但具有未定义的值，请使用未定义的值而不是删除运算符。在下面的示例中，trees[3] 被赋值为 undefined，但数组元素仍然存在:

var trees = ["redwood","bay","cedar","oak","maple"];
trees[3] = undefined;
if (3 in trees) {
// this gets executed
}

Answer 1

考虑一下:

var a = [1, 2, 3];
a[1] = undefined;
console.log('1' in a); // true
delete a[1];
console.log('1' in a); // false
console.log(a.length); // 3
console.log(a); // [1, undefined, 3]

如果你想在数组中间删除你应该使用拼接:

var a = [1, 2, 3];
a.splice(1, 1);
console.log(a.length); // 2
console.log(a); // [1, 3]

拼接的参数是:

array.splice(start, deleteCount[, item1[, item2[, ...]]])

See more over at MDN