Function call during loop is incorrectly returning undefined

Question

New to js and I'm attempting to solve a really simple problem. I'm not sure why it's not working though. Must be a js quirk I am unfamiliar with. Can someone tell me why x would return back as undefined thus causing my function to return false when it should return true? I'm trying to replicate the 'every' method for arrays and return false if one of the array elements returns false from my callback.

I've attempted to debug with webstorm and I still could not find the solution. Here is my code,

function every_loop(array, test) {
  for (let index = 0; array.length - 1; index++) {
    let x = test(array[index]);
    if (!x) {
      console.log(array[index], index);
      return false;
    }
  }
  return true;
}

console.log(every_loop([1, 2, 3, 4, 5], n => n >= 1));

My output is false when it should be true. Also, right before it outputs false, it shows undefined as a value for array[index] which leads me to believe that my for loop parameters are incorrect but that isn't the case either. Any help would be appreciated. Thanks

Well where's the test() code? Also your for loop test condition is wrong. edit oh wait I see, sorry — Pointy
– Pointy, Commented Jan 24, 2019 at 16:42
@Pointy Yea for some reason, no matter what I change the 2nd parameter in for loop to, say array.length -2, it still goes through the array one more time than it should and tries to find a fifth index. this is undefined. I don't know why it is doing that though — UCProgrammer
– UCProgrammer, Commented Jan 24, 2019 at 16:45
The second expression has to be a comparison, not just a number. — Pointy
– Pointy, Commented Jan 24, 2019 at 16:46
@UCProgrammer You would notice the problem if you log index as well. It is getting out of bound. — Yury Tarabanko
– Yury Tarabanko, Commented Jan 24, 2019 at 16:47

Prince Hernandez · Accepted Answer · 2019-01-24 16:45:41Z

2

you are missing the clause of the second parameter of the for loop. changing it to index <= array.length - 1 fixes your code.

function every_loop(array, test) {
  for (let index = 0; index <= array.length - 1; index++) {
    let x = test(array[index]);
    if (!x) {
      console.log(array[index]);
      return false;
    }
  }
  return true;
}

console.log(every_loop([1, 2, 3, 4, 5], n => n >= 1));

answered Jan 24, 2019 at 16:45

Prince Hernandez

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2 Comments

UCProgrammer Over a year ago

jeez louis. thanks. Should I delete this question? I doubt anyone would be as stupid as me and forget that. Been answered tons of times before I assume

FZs Over a year ago

@UCProgrammer See this

Mulan · Accepted Answer · 2019-01-24 16:46:31Z

1

Your for loop is broken. I would recommend for..of -

function everyLoop (arr, test) {
  for (const x of arr)
    if (!test(x))
      return false
  return true
}

console.log(everyLoop([1,2,3,4,5], n => n >= 1)) // true
console.log(everyLoop([1,2,3,4,5], n => n >= 2)) // false

You said it's practice, but just know JavaScript does include a built-in Array#every that does exactly this -

console.log([1,2,3,4,5].every(n => n >= 1)) // true
console.log([1,2,3,4,5].every(n => n >= 2)) // false

answered Jan 24, 2019 at 16:46

Mulan

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2 Comments

UCProgrammer Over a year ago

thanks and I'm aware. part of the problem was to do it both ways and did the solution with every method in a second's time and then switched to the loop and thought it would be simple as well. Little did I know I'd forget something as simple as a condition on my 2nd parameter in for loop and end up spending 20 minutes on the problem

Mulan Over a year ago

Try not to beat yourself up. JavaScript is a loose, multi-paradigm language making it possible to solve the same problem in countless ways. It's difficult for learners to find the important things to focus on. Other languages have better tooling that would make it easier to spot this bug, or make it altogether impossible to even write the bug (compile error).

FZs · Accepted Answer · 2019-01-24 16:48:17Z

1

Your condition in for loop will be always true. (array.length-1==4). So use this instead:

function every_loop(array, test) {
        for (let index = 0; index<array.length;index++) {
            let x = test(array[index]);
            if (!x) {
                return false;
            }
    }
    return true;
}

console.log(every_loop([1,2,3,4,5], n => n >= 1));
console.log(every_loop([0,1,2,3,4,5], n => n >= 1));

answered Jan 24, 2019 at 16:48

FZs

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Comments

Asim Khan · Accepted Answer · 2019-01-24 16:50:24Z

1

replace

for (let index = 0; array.length - 1; index++) {

with

for (let index = 0; index < array.length; index++) {

answered Jan 24, 2019 at 16:50

Asim Khan

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