68 lines
1.2 KiB
Markdown
68 lines
1.2 KiB
Markdown
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---
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id: 56104e9e514f539506016a5c
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title: 使用 For 循環遍歷數組的奇數
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challengeType: 1
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videoUrl: 'https://scrimba.com/c/cm8n7T9'
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forumTopicId: 18212
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dashedName: iterate-odd-numbers-with-a-for-loop
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---
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# --description--
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對於循環，一次不必遞增一個。 通過更改我們的 `final-expression`，我們可以用偶數來計數。
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初始化 `i = 0`，當 `i < 10` 的時候繼續循環。 `i += 2` 讓 `i` 每次循環之後增加 2。
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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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```
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循環結束後，`ourArray` 的值爲 `[0,2,4,6,8]`。 改變計數器（`initialization`） ，這樣我們可以用奇數來遞增。
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# --instructions--
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寫一個 `for` 循環，把從 1 到 9 的奇數添加到 `myArray`。
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# --hints--
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應該使用 `for` 循環。
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```js
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assert(/for\s*\([^)]+?\)/.test(code));
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```
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`myArray` 應該等於 `[1,3,5,7,9]`
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```js
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assert.deepEqual(myArray, [1, 3, 5, 7, 9]);
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```
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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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# --solutions--
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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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