Javascript function(to sum elemnts of array) dossn't work

Question

I make a function(in javascript) to sum the elements of an array and return the result ... but there is a problem and the output of this function is "NaN" but when i checked the type of variable , that was "number" :| please please help me to solve this ...

function arrSum(a) {
   var val = 0;
   for (var i = 0; i <= a.length; i++) {
      val += a[i];
   }
      return val;
}
var testArr = [1, 2, 3, 4];
var pass = arrSum(testArr);
document.write(pass);

// SHOW "NaN"

function arrSum(a) {
   var val = 0;
   for (var i = 0; i <= a.length; i++) {
      val += a[i];
   }
      return val;
}
var testArr = [1, 2, 3, 4];
var pass = arrSum(testArr);
document.write(typeof pass);

// SHOW "number"

i am sorry about my english. i am not good at it :) and thank you for your helps ...

Where is result defined?

user47589
– user47589

2018-03-07 14:31:10 +00:00
Commented Mar 7, 2018 at 14:31 — user47589
– user47589, Commented Mar 7, 2018 at 14:31

yajiv · Accepted Answer · 2018-03-07 15:46:11Z

3

Very much simple syntax(for array sum).

var test = [ 1, 2, 3, 4 ];
sum = eval( test.join("+") );  //join will return string which is '1+2+3+4' and eval will evaluate it and give result which is sum of numbers.
console.log(sum);

You are returning result but your sum is in val.

And also for loop condition is incorrect.

The reason you are getting NaN because at last iteration you are doing number+undefined which will result in NaN and type of NaN is number.

Hope it's clear your doubt why you are getting NaN and number.

Below is demo of correct code.

function arrSum(a) {
   var val = 0;
   for (var i = 0; i < a.length; i++) {
      val += a[i];
   }
      return val;
}
var testArr = [1, 2, 3, 4];
var pass = arrSum(testArr);
document.write(pass);

For additional info

Conversion rules

operand + operand = result

If at least one operand is an object, it is converted to a primitive value (string, number or boolean);
After conversion, if at least one operand is string type, the second operand is converted to string and the concatenation is executed;
In other case both operands are converted to numbers and arithmetic addition is executed.

If both operands are primitive types, then operator checks if at least one is string and executes the concatenation. In other case it just transforms everything to numbers and sum.

Example:-

var result = 12 + undefined; // NaN

Explanation:

12 + undefined (Because none of the operands is string, convert the undefined to a number NaN based on rule 3)
12 + NaN (Numbers addition)
NaN

edited Mar 7, 2018 at 15:46

answered Mar 7, 2018 at 14:33

yajiv

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2 Comments

georg Over a year ago

It would be helpful to explain them why they're getting NaN and why their typeof prints number

AmiR Over a year ago

but your explain was very helpful

gurvinder372 · Accepted Answer · 2018-03-07 14:33:28Z

2

For-loop's ending condition is incorrect, array's index is 0-based

Make it

for (var i = 0; i < a.length; i++) {

Also return val instead of result.

Demo

function arrSum(a) {
   var val = 0;
   for (var i = 0; i < a.length; i++) {
      val += a[i];
   }
      return val; //observe this line
}
var testArr = [1, 2, 3, 4];
var pass = arrSum(testArr);
console.log(pass);

edited Mar 7, 2018 at 14:33

answered Mar 7, 2018 at 14:31

gurvinder372

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2 Comments

user47589 Over a year ago

Also, the function returns an undefined variable.

gurvinder372 Over a year ago

@Amy yeah and that as well. Was just about to submit my edit :)

Basel Issmail · Accepted Answer · 2018-03-07 14:54:14Z

0

Another way to write arrSum function in one line with no loops using using reduce (ES5):

function arrSum(arr) {
 return arr.reduce((sum, x) => sum + x);
}

var testArr = [1, 2, 3, 4];
var pass = arrSum(testArr);
document.write(pass);

answered Mar 7, 2018 at 14:54

Basel Issmail

4,0657 gold badges24 silver badges37 bronze badges

Comments

user7803261 · Accepted Answer · 2018-03-07 14:33:06Z

0

The problems are return result and for (var i = 0; i <= a.length; i++). You must use respectively return val and for (var i = 0; i < a.length; i++).

answered Mar 7, 2018 at 14:33

user7803261

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