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const arr1 = [1, 2, 3, 4, 5];
const arr2 = [];
let index = Math.floor(Math.random() * arr1.length);

for(let i = 0; i < arr1.length; i++) {
    arr2.push(arr1[index]);
    do {
        index = Math.floor(Math.random() * arr1.length);
        console.log(arr2.includes(arr1[index])); //thís is to see the result
    } while (arr2.includes(arr1[index]))
}

console.log('arr2: ', arr2);

I want to make the new unordered array from the first array. But when I run this code, the loop becomes an infinite loop although the conditional is false. My question is the reason make the loop still running with the false conditional?

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Barmar
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Mai Linh
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    Your code will hang inside the last do-while loop forever since at that point, every element has been pushed to the new array. Shuffling an array doesn't require a check like that if you remove the element from the original array (or a copy of it). –  Sep 17 '21 at 16:03
  • Add a `console.log()` in the outer loop and you'll see that the inner loop is not running when the condition is false. – Barmar Sep 17 '21 at 16:09
  • Make a simple copy of the original array and then perform a Fisher-Yates shuffle. – Pointy Sep 17 '21 at 16:09
  • See https://stackoverflow.com/questions/2450954/how-to-randomize-shuffle-a-javascript-array for the correct ways to shuffle an array. – Barmar Sep 17 '21 at 16:10
  • I understand the problem. I will try again with the solution you gave me. Thanks you for all your help ☺️ – Mai Linh Sep 17 '21 at 16:23

0 Answers0