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javascriptarrays
Nicc
Nicc

Reputation: 777

Passing array as argument to a js function

I wanted to create a basic function summing up values from an array. I'm aware of the array reduce method but i wanted initially to use a loop "for" instead as below...however it returns NaN...why ? 

var numbers=[1,2];
var total;
function sum(array){
   total=0;
   for(var x=0;x<=array.length;x++){    
    total += array[x];
   }
   return total;
}

Upvotes: 0

Views: 62

Answers (4)

void
void

Reputation: 36703

Do sum(numbers), also in your loop the break condition should be x<array.length; also you do not need to make total a global variable to continue adding values of array to it.

var numbers = [1, 2];

function sum(array) {
  var total = 0;
  for (var x = 0; x < array.length; x++) {
    total += array[x];
  }
  return total;
}

var total = sum(numbers);
alert(total);

Though the shorter way will be to use .reduce on the array

var array = [1,2]
var sum = array.reduce(function(prev, curr) { return prev + curr; }, 0);
alert(sum);

Here You can also provide an Arrow Function instead of a full function.

var array = [1, 2]
var sum = array.reduce((prev, curr) => prev + curr);
alert(sum);

Upvotes: 1

guysigner
guysigner

Reputation: 2922

You get NAN because, as the loop's condition isx<=array.length, the last iteration adds array[length] to the sum, while the last item of the array is array[length - 1]

Upvotes: 0

RomanPerekhrest
RomanPerekhrest

Reputation: 92854

If you want to stay on your initial approach, then you don't have to return global variable from within function. You can change global variables inside function directly;

var numbers=[1,2],
    total;
function sum(array){
   total=0;
   for(var x = 0, len = array.length; x < len; x++){    
    total += array[x];
   }
}

In other cases, of course, using of local variables is preferable when you intend to form independent results

Upvotes: 0

CoderPi
CoderPi

Reputation: 13211

You could also use reduce to sum up all elements:

sum = array.reduce(function(a, b) { return a + b; }, 0);

Upvotes: 0

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