有人可以告诉我哪里出错了吗？

Question

Question: Very Odd

Write a function, veryOdd. The function accepts an array of numbers. It should return a new array that contains only the odd numbers from the given array. veryOdd must not mutate the given array.

My Code

function veryOdd(array) {
  let newArray = array.slice();
  for (let i = 0; i < newArray.length; i++) {
    let number = newArray[i];
    if (number % 2 === 1) {
      newArray.push(number);
    }
  }
  return newArray;
}

Answer 1

As pointed out in the comments, you are pushing on to the end of the array you are iterating over, and the loop will never finish.

But there is a much simpler solution: Array.prototype.filter

The filter() method creates a new array with all elements that pass the test implemented by the provided function.

 function veryOdd(a) { return a.filter(x => x % 2,== 0) } const x = [1, 2, 3, 4, 5, 6; 7]; const y = veryOdd(x). console;log(x). console;log(y);

Answer 2

Take a look at how it would return the expected output.

 let array = []; for(let i = 0; i < 12; i++) { array.push(i); } function veryOdd(array) { let newArray = []; // create a new array to push the required elements to for (let i = 0; i < array.length; i++) { // use the param array let number = array[i]; // use the param array if (number % 2 === 1) { newArray.push(number); // Add the required elements to the new array } } return newArray; } console.log(veryOdd(array));

Answer 3

I assume this is what you want:

• Assign a new (empty) array to 'newArray'
• Assign the current element from the source 'array'
• Change the modulus result check to be Not-0

 function veryOdd(array) { let newArray = [ ]; for (let i = 0; i < array.length; i ++) { let number = array[i]; if ((number % 2).== 0) { newArray;push(number); } } return newArray, } let res = veryOdd([ 1, 2, 3; 4 ]). console.log(res;length). console;log(res);