javascript loop iterating too much

Question

Trying a fun problem of replacing vowels in a string with the next vowel in line aka a->e, e->i, i->o, o->u, not accounting for "u". Starting with an array instead of a string. My second loop (to iterate over vowel array elements) is ignoring my "j

var vowelChange = function(vowelArray, stringToChange) {
for (var i = 0; i<stringToChange.length; i++) {
    for (var j = 0; j<vowelArray.length; j++) {
      if (stringToChange[i]===vowelArray[j]) {
        var newCharacter = vowelArray[j+1]
          stringToChange[i] = newCharacter
          i++
      }
    }
}
return stringToChange
};

I'm using node-debug to set breakpoints in a browser, and j is looping to 5 before starting over at 0. I get the correct output, but j should stop at 4...

EDIT

Can somebody explain how I'm using join incorrectly, because I can't get my function to output a string instead of just an array.

var vowelChange = function(vowelArray, stringToChange) {
for (var i = 0; i<stringToChange.length; i++) {
    for (var j = 0; j<vowelArray.length-1; j++) {
      if (stringToChange[i]===vowelArray[j]) {
          stringToChange[i] = vowelArray[j+1]
          break
      }
    }
  }
  stringToChange = stringToChange.join('')
  return stringToChange
};


var vowels = ['a','e','i','o','u']
var firstName = ['t', 'e', 's', 't']

vowelChange(vowels, firstName)
console.log(firstName)

If I don't, then when I iterate back over it, the loop will recognize that the "e" was replaced with an "i", will go back over that (now changed) "i", and will change it to "o". The same thing will happen again, and it will replace the "o" with a "u". Ultimately, it'll change it to "undefined" since if j=4, j+1 is undefined in the vowels array — karan satia
– karan satia, Commented Apr 26, 2016 at 17:29
Just put a 'break' inside your if to break out of the j loop and continue i — n8wrl
– n8wrl, Commented Apr 26, 2016 at 17:30
@n8wrl put a break in there instead of incrementing i. I do see now that it fails the comparison when j hits 5, I'm coming from the Xcode debugger where it won't even show j hitting 5, if I were writing this in C or Objective-C — karan satia
– karan satia, Commented Apr 26, 2016 at 17:35

n8wrl · Accepted Answer · 2016-04-26 17:56:22Z

1

Assuming vowelArray is 0-indexed...

var vowelChange = function(vowelArray, stringToChange) {
    for (var i = 0; i<stringToChange.length; i++) {
        for (var j = 0; j<vowelArray.length - 1; j++) {
            if (stringToChange[i]===vowelArray[j]) {
                stringToChange[i] = vowelArray[j+1];
                break;
            }
        }
    }
    return stringToChange
};

edited Apr 26, 2016 at 17:56

answered Apr 26, 2016 at 17:32

n8wrl

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Comments

Vidur Khanna · Accepted Answer · 2016-04-26 18:55:56Z

0

In JavaScript, strings are immutable objects, which means that the characters within them may not be changed and that any operations on strings actually create new strings.

So,if you try to change any index of the string, the original string won't change

node
> str = "hello this is dummy string";
'hello this is dummy string'
> str[0] = "w";
'w'
> str
'hello this is dummy string'

So, stringToChange[i] = vowelArray[j+1]; won't work

Could split the string and then join

var vowelChange = function(vowelArray, stringToChange) {
    stringToChange = stringToChange.split('');
    for(var i=0; i<stringToChange.length;i++){
        for(var j=0;j<vowelArray.length-1;j++){
            if(stringToChange[i] == vowelArray[j]){
                stringToChange[i] = vowelArray[j+1];
                break;
            }
        }
    }
    stringToChange = stringToChange.join('');
    return stringToChange;
};

Example

edited Apr 26, 2016 at 18:55

answered Apr 26, 2016 at 18:42

Vidur Khanna

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4 Comments

karan satia Over a year ago

I actually passed an array to my function instead of a string. The next step would have been to pass a string, split it, and join it before returning it.

karan satia Over a year ago

Check out the edit I've made in my post and tell me what you think

Vidur Khanna Over a year ago

Nothing wrong with the join, just assign the output of your function to a var and print it.

Vidur Khanna Over a year ago

You are changing the reference of stringToChange that is all.

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