[英]Why do these ways of returning arrays differ in Javascript?
I have two examples: 我有两个例子:
function myFunction(arr, size) {
// Break it up.
var newArray = [];
var tempArray = arr.splice(0,1);
newArray.push(arr.splice(0,1));
console.log(newArray);
return arr;
}
myFunction(["a", "b", "c", "d"], 2);
And this one: 还有这个:
function myFunction(arr, size) {
// Break it up.
var newArray = [];
var tempArray = arr.splice(0,1);
newArray.push(tempArray);
console.log(newArray);
return arr;
}
myFunction(["a", "b", "c", "d"], 2);
The main difference being the line: 主要区别在于:
newArray.push(arr.splice(0,1)); // Or newArray.push(tempArray);
Why does the first example return Array[1] and the second example return ["a"]? 为什么第一个示例返回Array [1],第二个示例为何返回[“ a”]?
I was expecting to get ["a"] regardless of which way I went, can someone possibly help me understand what is happening here? 无论我走哪条路,我都希望得到[“ a”],有人可以帮助我了解这里发生了什么吗?
I was just trying to take the first element of the array by splicing (which I believe returns an array of the removed elements) and push this array onto my "newArray" so I can ultimately have an array containing nested arrays holding each character. 我只是想通过拼接来获取数组的第一个元素(我相信该元素会返回已删除元素的数组),然后将该数组推入我的“ newArray”,以便最终得到一个包含嵌套数组的数组,其中每个数组都包含嵌套数组。 Ie [["a"],["b"],["c"]...]
即[[“ a”],[“ b”],[“ c”] ...]
EDIT: Please ignore the "size" parameter. 编辑:请忽略“大小”参数。 EDIT2: Sincere apologies.
EDIT2:真诚的道歉。 I'm not worried about the return statement.
我不担心退货声明。 Stupid of me to forgot to mention.
我笨到忘了提。 I'm looking at my console.log output.
我正在查看我的console.log输出。 When I run the script and look at the console, that's when I'm getting Array[1] or ["a"].
当我运行脚本并查看控制台时,就是在获取Array [1]或[“ a”]的时候。
This is the most important information you need to have in mind: splice
changes the original array (read this: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/splice ). 这是您需要记住的最重要的信息:
splice
更改了原始数组(请阅读: https : //developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/splice )。
So, let's see your first function. 因此,让我们看看您的第一个功能。 This line:
这行:
var tempArray = arr.splice(0,1);
Changes arr
, as expected. 按预期更改
arr
。 But, when you do this in the next line: 但是,当您在下一行中执行此操作时:
newArray.push(arr.splice(0,1));
You are changing arr
again! 您再次更改了
arr
!
Your second function, on the other hand, changes arr
just once: 另一方面,您的第二个功能仅更改
arr
一次:
var tempArray = arr.splice(0,1);
newArray.push(tempArray);
In a nutshell: just count how many times you have a splice
in your first function (2 times, changing the original array twice) and how many times you have a splice
in your second function (1 time, changing the original array only once). 简而言之:只需计算您在第一个函数中有一个
splice
次数(2次,将原始数组更改两次)和在第二个函数中有一个splice
次数(1次,仅更改一次原始数组) 。
EDIT: regarding the console.log, newArray
is an array with an array, and that's why sometimes you see Array[1]
. 编辑:关于console.log,
newArray
是带有数组的数组,这就是为什么有时您会看到Array[1]
。 Do this: 做这个:
console.log(newArray[0]);
And now you'll see ["a"]
, ["b"]
or whatever. 现在,您会看到
["a"]
, ["b"]
或其他内容。
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