CODEWITHSUNDEEP
javascript
user14345364
user14345364

Reputation:

Array inside function returns undefined

I practice simple js lessons and have problem with this part below Function should return array index 0 instead of letter A but actually returns - undefined And i cant figure it why ?

function parseGuesses(guess) {
    var alphabet = ["A", "B", "C", "D", "E", "F", "G"];
    if (guess === null || guess.length !== 2) {
        alert("Ups proszę podać literę i cyfrę");
    } else { firstChar = guess.charAt(0);
        var row = alphabet.indexOf(firstChar);
    }
}
console.log(parseGuesses("A2"));

Upvotes: 0

Views: 71

Answers (2)

German
German

Reputation: 1166

It logs undefined because your function does not return anything.

Your function should have return keyword typically used at last line of the function like this.

function parseGuesses(guess) {

  // operation of something ....

  return something
}

Then logs it to the console

console.log(parseGuesses("A2"))

Modified from you code

 function parseGuesses(guess) {
  var alphabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G'];
  let firstChar, row;
  if (!guess || guess.length > 2) {
    alert('Ups proszę podać literę i cyfrę');
  } else {
    firstChar = guess.charAt(0);
    row = alphabet.indexOf(firstChar);
  }
  return [firstChar, row]; // return an array after those operation above
}
console.log(parseGuesses('A2'));

Upvotes: 0

Saar Davidson
Saar Davidson

Reputation: 1382

As Ivar said the function you wrote isn't returning anything so by default it would return undefine so to change the return value of the function you should write it like so:

function parseGuesses(guess) {
  var alphabet = ["A", "B", "C", "D", "E", "F", "G"];
  if (guess === null || guess.length !== 2) {
    alert("Ups proszę podać literę i cyfrę");
  } else {
    let firstChar = guess.charAt(0); // you should also declare when you create a new variable
    return alphabet.indexOf(firstChar);
  }
}

Upvotes: 1

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