freeCodeCamp/curriculum/challenges/korean/02-javascript-algorithms-and-data-structures/basic-javascript/nesting-for-loops.md

id, title, challengeType, videoUrl, forumTopicId, dashedName

id	title	challengeType	videoUrl	forumTopicId	dashedName
56533eb9ac21ba0edf2244e1	Nesting For Loops	1	https://scrimba.com/c/cRn6GHM	18248	nesting-for-loops

--description--

If you have a multi-dimensional array, you can use the same logic as the prior waypoint to loop through both the array and any sub-arrays. 여기 예시가 있습니다.

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

for (let i = 0; i < arr.length; i++) {
  for (let j = 0; j < arr[i].length; j++) {
    console.log(arr[i][j]);
  }
}

This outputs each sub-element in arr one at a time. Note that for the inner loop, we are checking the .length of arr[i], since arr[i] is itself an array.

--instructions--

Modify function multiplyAll so that it returns the product of all the numbers in the sub-arrays of arr.

--hints--

multiplyAll([[1], [2], [3]]) should return 6

assert(multiplyAll([[1], [2], [3]]) === 6);

multiplyAll([[1, 2], [3, 4], [5, 6, 7]]) should return 5040

assert(
  multiplyAll([
    [1, 2],
    [3, 4],
    [5, 6, 7]
  ]) === 5040
);

multiplyAll([[5, 1], [0.2, 4, 0.5], [3, 9]]) should return 54

assert(
  multiplyAll([
    [5, 1],
    [0.2, 4, 0.5],
    [3, 9]
  ]) === 54
);

--seed--

--seed-contents--

function multiplyAll(arr) {
  let product = 1;
  // Only change code below this line

  // Only change code above this line
  return product;
}

multiplyAll([[1, 2], [3, 4], [5, 6, 7]]);

--solutions--

function multiplyAll(arr) {
  let product = 1;
  for (let i = 0; i < arr.length; i++) {
    for (let j = 0; j < arr[i].length; j++) {
      product *= arr[i][j];
    }
  }
  return product;
}