了解JavaScript中的嵌套循环 [英] Understanding nested for loops in javascript

问题描述

此刻我正在freecodecamp上学习JavaScript，他们在其中的一个练习中有一个嵌套循环的示例:

I'm learning JavaScript at the moment on freecodecamp and they have an example for nested for loops in one of their excercises:

 var arr = [[1,2], [3,4], [5,6]];
 for (var i=0; i < arr.length; i++) {
  for (var j=0; j < arr[i].length; j++) {
    console.log(arr[i][j]);
  }
}

使用console.log = 1 2 3 4 5 6未定义.

With console.log = 1 2 3 4 5 6 undefined.

我或多或少了解for循环，并且我了解[i]和[j]用于访问数组(我认为吗?).我只是不明白为什么最后只打印出这些数字?我发现这个问题是在几年前问的，但这只是解释了如何编写它们，而不是它们是如何工作的:

I understand for loops more or less, and I understand that [i] and [j] are used to access the array (I think?). I just don't understand why at the end it just prints out those numbers? I found this question asked a few years back but it just explains how to write them, not how they work:

用于多维javascript数组中的循环

我将其分解为:

var arr = [  [1,2], [3,4], [5,6]];for (var i=0; i < arr.length; i++) {
 console.log(arr[i]);}

哪个打印出

[ 1, 2 ]
[ 3, 4 ]
[ 5, 6 ]
undefined

和

var arr = [  [1,2], [3,4], [5,6]];
for (var i=0; i < arr.length; i++) {  
 for (var j=0; j < arr[i].length; j++) {    
  console.log(arr[i]);  }}

打印出的内容:

[ 1, 2 ]
[ 1, 2 ]
[ 3, 4 ]
[ 3, 4 ]
[ 5, 6 ]
[ 5, 6 ]
undefined

和

var arr = [  [1,2], [3,4], [5,6]];
 for (var i=0; i < arr.length; i++) {  
  for (var j=0; j < arr[i].length; j++) {    
   console.log(arr[j]);  }}

打印出来的

[ 1, 2 ]
[ 3, 4 ]
[ 1, 2 ]
[ 3, 4 ]
[ 1, 2 ]
[ 3, 4 ]
undefined

我了解前两个arr [i].循环遍历数组，并打印出各个元素(在本例中为数组)，在第二个中，我猜它只做了两次，因为有两个循环.我不明白的是:

I understand the first two arr[i]. The loop iterates through the array and prints out the individual elements (in this case an array) and in the second one I guess it just does it twice because there are two loops. What I don't understand is:

1. 为什么没有打印出arr [j]中的最后一个数组( [5，6]去吗?)
2. 为什么arr [i] [j]突然消除了数组而只是 打印出数字
3. 未定义"来自哪里

1. why the last array in arr[j] is not printed out (where did the [5, 6] go?)
2. why arr[i][j] suddenly eliminates the arrays and just prints out the numbers
3. where the 'undefined' comes from

有人可以帮助我解决这个问题，并解释在控制台中将其打印出来之前代码所采取的步骤吗?我真的很想了解它，但是甚至都不知道如何以正确的方式搜索这个问题.

Could anyone help me out with this and explain the steps the code takes before printing it out in the console? I would really like to understand it but don't even know how to search for this question the right way.

推荐答案

var arr = [[1,2], [3,4], [5,6]];

这是一个数组数组.像这样阅读起来会更容易一些:

This is an array of arrays. It is a little bit easier to read like this:

var arr = [
            [1,2],
            [3,4],
            [5,6]
          ];

这样一来，您可以轻松地看到一个由3个数组组成的数组.外部的"for"将循环遍历每个第一级数组.因此，当i = 0时，第一个外部for循环将获取第一个内部数组[1,2]:

That makes it a little bit easier to see that you have an array of 3 arrays. The outer 'for' will loop through each of 1st level arrays. So the very first outer for loop when i=0 you are going to grab the first inner array [1,2]:

for (var i=0; i < arr.length; i++) {
    //First time through i=0 so arr[i]=[1,2];
}

在内部循环中，您将一次循环遍历3个内部阵列中的每一个.

In the inner loop you are going to loop through each of the 3 inner arrays one at a time.

for (var j=0; j < arr[i].length; j++) {
    //Handle inner array.
}

此参数获取内部数组的长度:

This argument grabs the length of the inner array:

arr[i].length

因此，第一次通过外循环i = 0时，arr [i]等于[1,2]，因为您要获取第0个元素.请记住，数组元素总是从0开始计数，而不是从1开始.

So on your first time through the outer loop i=0 and arr[i] is going to equal [1,2] because you are grabbing the 0th element. Remember, arrays elements are always counted starting at 0, not 1.

最后，您将使用以下命令打印结果:

Finally you are printing out the results with:

console.log(arr[i][j]);

第一次，您可以将其分解. i = 0和j = 0. arr [0] [0]转换为从外部数组中获取第一个元素，然后从第一内部数组中获取第一个元素.在这种情况下，它是"1":

The first time through you can break it down a little. i=0 and j=0. arr[0][0] which translates as grab the first element from the outer array and then the first element from the first inner array. In this case it is '1':

[
    [1,2], <-- 0
    [3,4], <-- 1
    [5,6]  <-- 2
];

该代码将循环遍历第一个第一组[1,2]，然后遍历第二个[3,4]，依此类推.

The code will loop through the first first set [1,2], then the second [3,4], and so on.

这篇关于了解JavaScript中的嵌套循环的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！