Why does this JavaScript switch statement (which is inside a for loop) keep it's values from a previous iteration?

Question

function convertToRoman(num) {
  //seperate the number in to singular digits which are strings and pass to array.
  var array = ("" + num).split(""), 
      arrayLength = array.length, 
      romStr = "";

  //Convert the strings in the array to numbers
  array = array.map(Number);

  //Itterate over every number in the array
  for (var i = 0; i < array.length; i++) {

    //Calculate what power of ten the digit is by minusing it's index from the array length and passing it to variable "tenToPowerOf"
    var tenToPowerOf = arrayLength - array.indexOf(array[i]) - 1,
        low = "",
        mid = "",
        upp = "";

    //Set the low, mid and high variables based on their tenToPowerOf value
    switch (tenToPowerOf) {
    case 0:
      low = "I";
      mid = "V";
      upp = "X";
      break;
    case 1:
      low = "X";
      mid = "L";
      upp = "C";
      break;
    case 2:
      low = "C";
      mid = "D";
      upp = "M";
      break;
    case 3:
      low = "M";
      mid = "¯V";
      upp = "¯X";
      break;
    }

    //Check for digit value and add it's Roman Numeral value (based on it's power from the above switch statement) to romStr
    //The problem is, switch "remembers" the value of low, mid and high from the last time it itterated over the number. Thus 99 becomes "XCXC" and not "XCIX". 
    switch (array[i]) {
    case 1:
      romStr += low;
      break;
    case 2:
      romStr += low.repeat(2);
      break;
    case 3:
      romStr += low.repeat(3);
      break;
    case 4:
      romStr += low + mid;
      break;
    case 5:
      romStr += mid;
      break;
    case 6:
      romStr += mid + low;
      break;
    case 7:
      romStr += mid + low.repeat(2);
      break;
    case 8:
      romStr += mid + low.repeat(3);
      break;
    case 9:
      romStr += low + upp;
      break;
    case 10:
      romStr += upp;
      break;
    default:
      break;

    }

  }
  //return array;
  return romStr;
}

convertToRoman(99);

Yes, I have spent my share share of time looking for a relevant answer before asking (1.5+ hours).
The function converts numbers in to Roman Numerals.
- It takes a number argument passed to it
- Splits the number in to digits
- Converts then in to an array
- Then based on the number's power of ten (or position relevant to the length of the number) sets the low mid and high values
- These values are then applied in the order dictated by Roman Numerals and pushed to a string
- Finally, the string is returned.
It works with numbers that have digits occurring no more than once. The reason for this is because when the switch case is matched more than once in the for loop, it applies the low, mid and high values from the previous iteration.
The question is in the title but I would also like to ask for a solution to the problem that I am trying to solve please.
I will gladly provide more information and will answer every question

Can you please put relevant code in your question along with your actual question? (links die) — Naftali, Apr 08 '16 at 14:46

score 1 · Accepted Answer · answered Apr 08 '16 at 14:55

1

It seems to me your problem is in the use of array.indexOf(array[i]) to calculate the power. Guess what, if you have the same value in your array twice, the first found index is returned, not the index of your current element.

Guess what the index of your current element actually is? → i
No need for indexOf.

Nothing to do with switch.

answered Apr 08 '16 at 14:55

deceze

510,633
85
743
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Thank you for answering and editing my question. This solved my problem. – Tom Apr 09 '16 at 11:19

score 0 · Answer 2 · answered Apr 08 '16 at 14:47

0

Because Javascript uses function closures and your loop doesn't reset the values by default, in other words the variables inside the for still exists outside of it.

answered Apr 08 '16 at 14:47

Dellirium

1,362
16
30

score 0 · Answer 3 · answered Apr 08 '16 at 14:51

Variables declared as var inside a for loop are not reset on each iteration, it has nothing to do with the switch.

Try pasting this into your console -

for (var i = 1; i <= 10; i++)
{
    console.log('before', i, j);
    var j = i * 10;
    console.log('after', i, j);
}

Note that on the first loop, the "before" j is undefined, and then after that is always one step behind. If you run the same code again, j will start at 100

To fix this, I would set j to null (or some other sensible value) at the start of the loop -

for (var i = 1; i <= 10; i++)
{
    var j = null;
    console.log('before', i, j);
    j = i * 10;
    console.log('after', i, j);
}

Why does this JavaScript switch statement (which is inside a for loop) keep it's values from a previous iteration?

3 Answers3