Why is this result [undefined, undefined, undefined] and not [ ]?

I have questions
1 
const result = [1,2,3].map(n => {
  if(n === 0) {
     return 'hi'
  }
})

console.log(result) // [undefined, undefined, undefined]
2

When a function exits without using a return, by default, it'll return undefined.

const fn = () => {};
console.log(fn()); // undefined

So, your code is the same as this:

const result = [1,2,3].map(n => {
  if(n === 0) {
     return 'hi'
  } else {
    return undefined
  }
})

console.log(result) // [undefined, undefined, undefined]

And, .map() doesn't give any special treatment to the undefined value - if that gets returned, it'll stick it in the array as normal.

If you're wanting to also filter values out as you map, you'd either use a separate .filter(...) first, or you use .flatMap() like this:

const result = [1,2,3].flatMap(n => {
  if(n === 0) {
     return ['hi']
  } else {
    return []
  }
})

console.log(result) // []
3

Because the array doesn't contain 0, it only contains 1, 2, and 3.

Array iteration callbacks take the value as the first parameter, and the index as the second.

4

Wow ... ! Perfect solution!

const objectArray = [{id: 1, title: 'Hello'}, {id: 2, title: 'Hello'}, {id: 3, title: 'Hello'}];

// Before
const result = objectArray.flatMap(n => {
  if (n.id > 3) {
     return n.id
  }
})
console.log(result); // [undefined, undefined, undefined]

// After
const result = objectArray.flatMap(n => {
  if (n.id > 3) {
     return [n.id]
  } else {
     return []
  }
})
console.log(result); // []