--- id: 56533eb9ac21ba0edf2244e1 title: Nesting For Loops challengeType: 1 videoUrl: '' localeTitle: 嵌套循环 --- ## Description
如果您有一个多维数组,则可以使用与先前路点相同的逻辑来遍历数组和任何子数组。这是一个例子:
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 ++){
的console.log(ARR [i] [j]);
}
}
这个输出在每个子元件arr一次一个。注意,对于内部循环,我们检查arr[i].length ,因为arr[i]本身就是一个数组。
## Instructions
修改函数multiplyAll ,使其乘以product变量乘以arr的子数组中的每个数字
## Tests
```yml tests: - text: 'multiplyAll([[1],[2],[3]])应该返回6' testString: assert(multiplyAll([[1],[2],[3]]) === 6); - text: 'multiplyAll([[1,2],[3,4],[5,6,7]])应返回5040' testString: assert(multiplyAll([[1,2],[3,4],[5,6,7]]) === 5040); - text: 'multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]])应该返回54' testString: assert(multiplyAll([[5,1],[0.2, 4, 0.5],[3, 9]]) === 54); ```
## Challenge Seed
```js function multiplyAll(arr) { var product = 1; // Only change code below this line // Only change code above this line return product; } // Modify values below to test your code multiplyAll([[1,2],[3,4],[5,6,7]]); ```
## Solution
```js // solution required ```