Simple Adding square of digits of a number

Question

The result of this script is not the sum of the squares of the digits of a number, the first function works fine, I guess the problem is in the second

function nToArray(a) {
  var res=[];
  var s =a.toString();
  for(i=0;i<s.length;i++){
    res[i]=s.charAt(i);}
  return res;
}

 //var a ="hola";
 //document.write(a.length);
alert(nToArray(562));
function addSq(b){
  var c=nToArray(b);
  var z=0;
  var i;
  for (i=0;i<=c.length;i++){//here 
    z+=((parseInt(c[i]))^2);}
  return z;
}

alert(addSq(81));

For 81 I get 15, I don't get it.

`^` is the bitwise XOR operator. You are looking for `Math.pow` — MinusFour, Jan 03 '16 at 18:49
@MinusFour thank you, as you can see in starting to learn programming :S — Kepol, Jan 03 '16 at 18:50
Please take a look at http://ericlippert.com/2014/03/05/how-to-debug-small-programs/. — , Jan 03 '16 at 18:51
At some point in the future (ES2016?), you should be able to use `**` for exponentation. See https://github.com/rwaldron/exponentiation-operator. Actually, you can already use it in Babel. — , Jan 03 '16 at 19:00
See also http://stackoverflow.com/questions/30873116/how-do-i-square-a-numbers-digits. — , Jan 03 '16 at 19:03

T.J. Crowder · Answer 1 · 2016-01-03T18:52:45.700

2

JavaScript doesn't have an exponentiation operator, ^ is a bitwise integer XOR.

You're looking for Math.pow.

z += Math.pow(parseInt(c[i]), 2);

or of course, just use * to multiply:

var value = parseInt(c[i]);
z += value * value;

edited Jan 03 '16 at 18:52

answered Jan 03 '16 at 18:51

T.J. Crowder

1,031,962
187
1,923
1,875

Now it tells me Nan... I'll check it out – Kepol Jan 03 '16 at 18:58
1

In your loop, you have `i<=c.length` so, last element is `NaN`, just use `<` instead of `<=` – MinusFour Jan 03 '16 at 19:02
Great, just teaching myself programming, this website is helping me a lot thank you! – Kepol Jan 03 '16 at 19:08
You fixed it, I knew there was a problem with the lenght – Kepol Jan 03 '16 at 19:09
Consider using spilt, map and reduce - it is shorter and will save you from of by one errors. – mzedeler Jan 03 '16 at 19:12

score 2 · Answer 2 · answered Jan 03 '16 at 18:57

2

Here is a short version - sorry about the missing formatting:

var a=81;
console.log(a.toString().split('').map(function(i) {return Math.pow(i, 2)}).reduce(function(a, b) {return a + b}));

answered Jan 03 '16 at 18:57

mzedeler

4,177
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score 1 · Answer 3 · answered Jan 03 '16 at 18:51

1

While ^ represents an exponent in some languages, that is not the case in JavaScript.

Change

z+=((parseInt(c[i]))^2);

to

var tmp = parseInt(c[i]);
z+= tmp * tmp;

answered Jan 03 '16 at 18:51

Eric J.

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guysigner · Answer 4 · 2016-01-04T08:03:39.393

1

Hope this helps:

EDIT

function addSq(num){
var numNoStr = parseInt(num, 10);
var z = 0;
while (numNoStr > 0) {
    var digit = numNoStr % 10;
    z += Math.pow(digit, 2);
    numNoStr = parseInt(numNoStr / 10 , 10);
}
return z;
}
alert(addSq(81));

edited Jan 04 '16 at 08:03

answered Jan 03 '16 at 19:01

guysigner

2,822
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You're not using `numNoStr`. – Jan 03 '16 at 19:07
That's correct. Should've (but forgot) replaced all inner-used "num" with "numNoStr", just in case the input is a string, my bad – guysigner Jan 03 '16 at 19:12

score 1 · Answer 5 · 2016-01-03T19:27:04.570

1

You can do this numerically:

function sum_of_squares_of_digits(n) {
  var s = 0;
  while (n) {
    var digit = n % 10;
    s += digit * digit;
    n = (n - digit) / 10;
  }
  return s;
}

Just for fun, try using regexp:

function sum_of_squares_of_digits(n) {
  var s = 0;
  n.toString().replace(/\d/g, d => s += d * d);
  return s;
}

edited Jan 03 '16 at 19:27

answered Jan 03 '16 at 19:06

Simple Adding square of digits of a number

5 Answers5