Here you can find the source of sum()
Array.prototype.sum = function() { var sum = 0;/*from www .java2s.com*/ for(var i = this.length; -- i >= 0 ;) { sum += (this[i] === null || this[i] === undefined) ? 0 : this[i]; } return sum; }
Array.prototype.sum = function () { var total = 0; var i = this.length; while (i--) { total += this[i]; return total; function isEmpty(obj) { ...
Array.prototype.sum = function () { return this.reduce((sum, n) => sum + Number(n), 0)
Array.prototype.sum = function () { return this.reduce(function (total, aValue) { return total + Number(aValue); }); };
var log = console.log.bind(console) Array.prototype.sum = function() { return this.reduce(function(previous, current) { return previous + current }) var numbers = [1, 2, 3, 4, 5, 6] var result = numbers.sum() log(result) ...
Array.prototype.sum=function(){ return this.reduce(function(s,n){return s+n})
Array.prototype.sum = function() { for(var sum = i = 0; i < this.length; i++) { sum += this[i] return sum var arr = [1,21,3,4,22,45,6,7,32]; console.log(arr.join("+")+"="+arr.sum());
Array.prototype.sum = function() { var i, sum; sum = 0; for (i=0; i<this.length; i++) { sum+=this[i]; return sum; };
Array.prototype.sum = function(){ for (var i = 0, sum = 0 ; i != this.length; ++i) { var n = this[i]; if(!isNaN(n)) { sum += n*1; return sum; }; ...
Array.prototype.sum = function(){ return this.reduce( (a,b) => a + b, 0)