闭包及相关题目

闭包，即函数连同它作用域链上的要找的这个变量，共同构成闭包。目的为暂存封装数据。如下为典型闭包案例：

function car(){
  var speed = 0
  function fn(){
    speed++
    console.log(speed)
  }
  return fn
}
var speedUp = car()
speedUp()   //1
speedUp()   //2

1.如下代码输出多少？如果想输出3，那如何改造代码？

var fnArr = [];
for (var i = 0; i < 10; i ++) {
  fnArr[i] =  function(){
    return i
  };
}
console.log( fnArr[3]() ) //输出10，返回函数但是未执行函数，for循环结束后输入fnArr[3],函数才执行，此时i等于10

思路：形成闭包，暂存数据

var fnArr[]
for(var i = 0, i<10, i++){
  fnArr[i]=(function(j){
    return function(){
       return j
  }
 })(i)
}
console.log( fnArr[3]() )

2.封装一个car对象，满足如下功能

var Car = (function(){
   var speed = 0;
   //补充
   return {
      setSpeed: setSpeed,
      getSpeed: getSpeed,
      speedUp: speedUp,
      speedDown: speedDown
   }
})()
Car.setSpeed(30)
Car.getSpeed() //30
Car.speedUp()
Car.getSpeed() //31
Car.speedDown()
Car.getSpeed()  //30

解答：

var Car = (function(){
   var speed = 0;
  function setSpeed(s){
    speed=s
  }
  function getSpeed(){
    return speed
  }
  function speedUp(){
    speed++
  }
  function speedDown(){
    speed--
}
   //补充
   return {
      setSpeed: setSpeed,
      getSpeed: getSpeed,
      speedUp: speedUp,
      speedDown: speedDown
   }
})()
Car.setSpeed(30)
Car.getSpeed() //30
Car.speedUp()
Car.getSpeed() //31
Car.speedDown()
Car.getSpeed()  //30

3.如下代码输出多少？如何连续输出 0,1,2,3,4？

for(var i=0; i<5; i++){
  setTimeout(function(){
    console.log('delayer:' + i )
  }, 0)
}

解答：
delayer:5 //setTimeout为异步处理，for循环结束后才开始执行，即i=5

for(var i=0;i<5;i++){
    satTimeout(function(j){
        return function(){
            console.log('delayer:'+ j)
        }
    })(i),0)
}  

4.补全代码，实现数组按姓名、年纪、任意字段排序。

var users = [
  { name: "John", age: 20, company: "Baidu" },
  { name: "Pete", age: 18, company: "Alibaba" },
  { name: "Ann", age: 19, company: "Tecent" }
]

users.sort(byField('age'))
users.sort(byField('company'))

解答：

function byName(user1, user2){
  return user1.name > user2.name
}

function byAge (user1, user2){
  return user1.age > user2.age
}
或者：
function byFeild(field){
  return function(user1, user2){
    return user1[field] > user2[field]
  }
}
users.sort(byField('company'))

5.写一个 sum 函数，实现如下调用方式。

console.log( sum(1)(2) ) // 3
console.log( sum(5)(-1) ) // 4
解答：

function cat(i){
  return function(j){
    return i+j
  }
}

参考：http://book.jirengu.com/fe/%E5%89%8D%E7%AB%AF%E5%9F%BA%E7%A1%80/Javascript/%E9%97%AD%E5%8C%85.html