2018-10-10 18:03:03 -04:00
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---
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id: 56104e9e514f539506016a5c
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2020-12-16 00:37:30 -07:00
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title: 使用 For 循环遍历数组的奇数
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2018-10-10 18:03:03 -04:00
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challengeType: 1
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2020-04-29 18:29:13 +08:00
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videoUrl: 'https://scrimba.com/c/cm8n7T9'
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forumTopicId: 18212
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2021-01-13 03:31:00 +01:00
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dashedName: iterate-odd-numbers-with-a-for-loop
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2018-10-10 18:03:03 -04:00
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---
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2020-12-16 00:37:30 -07:00
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# --description--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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for循环可以按照我们指定的顺序来迭代,通过更改我们的`计数器`,我们可以按照偶数顺序来迭代。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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初始化`i = 0`,当`i < 10`的时候继续循环。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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`i += 2`让`i`每次循环之后增加2。
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2018-10-10 18:03:03 -04:00
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```js
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var ourArray = [];
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for (var i = 0; i < 10; i += 2) {
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ourArray.push(i);
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}
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2020-12-16 00:37:30 -07:00
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```
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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循环结束后,`ourArray`的值为`[0,2,4,6,8]`。 改变`计数器`,这样我们可以用奇数来数。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --instructions--
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2020-04-29 18:29:13 +08:00
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2020-12-16 00:37:30 -07:00
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写一个`for`循环,把从 1 到 9 的奇数添加到`myArray`。
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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# --hints--
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2018-10-10 18:03:03 -04:00
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2020-12-16 00:37:30 -07:00
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你应该使用`for`循环。
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert(code.match(/for\s*\(/g).length > 1);
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2018-10-10 18:03:03 -04:00
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```
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2020-12-16 00:37:30 -07:00
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`myArray`应该等于`[1,3,5,7,9]`。
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2020-04-29 18:29:13 +08:00
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2018-10-10 18:03:03 -04:00
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```js
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2020-12-16 00:37:30 -07:00
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assert.deepEqual(myArray, [1, 3, 5, 7, 9]);
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2018-10-10 18:03:03 -04:00
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```
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2020-04-29 18:29:13 +08:00
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2021-01-13 03:31:00 +01:00
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# --seed--
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## --after-user-code--
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```js
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if(typeof myArray !== "undefined"){(function(){return myArray;})();}
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```
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## --seed-contents--
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```js
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// Setup
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var myArray = [];
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// Only change code below this line
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```
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2020-12-16 00:37:30 -07:00
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# --solutions--
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2021-01-13 03:31:00 +01:00
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```js
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var myArray = [];
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for (var i = 1; i < 10; i += 2) {
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myArray.push(i);
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}
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```
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