CODEWITHSUNDEEP
javascript
UCProgrammer
UCProgrammer

Reputation: 557

Function call during loop is incorrectly returning undefined

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

Upvotes: 0

Views: 60

Answers (4)

Asim Khan
Asim Khan

Reputation: 2049

replace 

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

with 

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

Upvotes: 1

FZs
FZs

Reputation: 18639

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));

Upvotes: 1

Mulan
Mulan

Reputation: 135415

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

Upvotes: 1

Prince Hernandez
Prince Hernandez

Reputation: 3731

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));

Upvotes: 2

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