nodejs 的async的BUG

let async = require('async');
let util  = require('../../util/utils');
test();

function test() {
    let task ={time:util.getLocalTime(),
    id:0};
    async.mapSeries([1,2,3],function (num, callback) {
        task.id = num;
        if(num>2){
            callback(null,{id:task.id,info:task});
        }else{
            callback(null,{id:task.id,info:task});
        }
    },function (err, results) {
        results.forEach(function (re) {
            console.log(re);
        })
    })
}

//打印的结果是：

//{ id: 1, info: { time: '2017-07-21 14:22:34', id: 3 } }
//{ id: 2, info: { time: '2017-07-21 14:22:34', id: 3 } }
//{ id: 3, info: { time: '2017-07-21 14:22:34', id: 3 } }

//因为我们把task定义在async外面，所以每一次返回的task是最后的task值。要规避此bug，我们直接把task的赋值在async里面就可以了