reduce()方法对累加器和数组中的每个元素（从左到右）应用一个函数，将其减少为单个值。
reduce接收两个参数，第一个function，第二个为初始值
//求数组中奇数的和
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce(function(x,y){
if(y%2===1){
return x+y //return 的x+y的值就赋给了function中的参数x
}
return x
},0)
//求数组中奇数的和
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce((x, y) => { return (y % 2 === 1) ? x + y : x; }, 0)
//求数组成员的和
var a = [1,2,3,4,5,6,7,8,9]
a.reduce((num,now)=>{ return num+now},0)
//45
var a = [1,2,3,4,5,6,7,8,9]
a.reduce((num,now)=>(num+now),0)
//45
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce(function (x, y) {
if (y % 2 === 1) {
return x
}
console.log(x + 'xxxxxx')
console.log(y + 'y')
}, 0)
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce(function (x, y) {
console.log(x + y)
return x+y
}, 0)
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce(function (x, y) {
if (y % 2 === 1) {
return x + y
}
}, 0)
var a = [1, 2, 3, 4, 5, 6, 7, 8, 9]
a.reduce(function(x,y){
x= 5
if(y%2===1){
return x+y
}
return x
},0)
