How can I sum array index values?-CodePudding

I am new to Javascript and at the moment I'm learning how "arrays" are used.

In my code below I have 12 numbers held by an array variable. Next, the for loop is iterating over the indexes to check which values have 2 or more digits, the while-loop then summarizes the digits (e.g. value '130' at index 8, will be 1 3 0=4).

Final step..and also where I'm stuck:

I need to sum up all the "new" index values and return the result in a variable. With the numbers provided in the code, the result would be '50'.

Anyone have clue on how to do this? I've tried the conventional for-loop with sum = array[i], but it doesn't work.

var arrChars = [4, 2, 14, 9, 0, 8, 2, 4, 130, 65, 0, 1];

for (var i = 0; i < arrChars.length; i  ) {
  var digsum = 0;

  while (arrChars[i] > 0) {
    digsum  = arrChars[i] % 10;

    arrChars[i] = Math.floor(arrChars[i] / 10);
  }

  var sum = 0; // this last part won't work and I just get "nan", 12 times

  for (var j = 0; j < arrChars.length; j  ) {
    sum  = parseInt(digsum[j]);
  }

  console.log(sum); // desired output should be '50'
}

CodePudding user response：

Move digsum outside and it will contain the sum of every number in it:

var arrChars = [4, 2, 14, 9, 0, 8, 2, 4, 130, 65, 0, 1];

var digsum = 0;
for (var i = 0; i < arrChars.length; i  ) {

  while (arrChars[i] > 0) {
    digsum  = arrChars[i] % 10;

    arrChars[i] = Math.floor(arrChars[i] / 10);
  }
}
console.log(digsum); // desired output should be '50'

CodePudding user response：

I'd make this easy and just flatten the array of numbers into a string of digits, split that into an array of single digits, and add them together:

var arrChars = [4, 2, 14, 9, 0, 8, 2, 4, 130, 65, 0, 1];

console.log([...arrChars.join('')].reduce((agg, cur) => agg  =  cur, 0));

CodePudding user response：

digsum is a number but you're trying to access it as an array (digsum[j]). You probably want to overwrite the array with digsum's value for each index, e.g. after the while block: arrChars[i] = digsum;