对于这三个概念经常混淆，它们的作用也不甚明了，总结一句话：
call()、apply()、bind() 都是用来重定义 this 这个对象的！
**
从示例看起

示例1：

const obj = {
  age:25,
  name:"Tom",
  fun1(){
    console.log("name:" + this.age + "\nage:" + this.name)
  }
}
const fun2 = obj.fun1
fun2()

输出结果：

name:undefined
age:undefined

分析原因：
fun()2 中的this指向了window

示例2：

var age = 26, name = "Jerry"
const obj = {
  age:25,
  name:"Tom",
  fun1(){
    console.log("name:" + this.age + "\nage:" + this.name)
  }
}
const fun2 = obj.fun1
fun2()

输出结果：

name:25
age:Tom

分析原因：

定义的age和name在全局作用域中，验证了示例1的分析原因

提出疑问，如果我还想让fun2中的this指向obj呢，那这么办，这时候就用到这三个关键字了

示例3

我们重新修改一下程序

var age = 26, name = "Jerry"
const obj = {
  age:25,
  name:"Tom",
  fun1(){
    console.log("name:" + this.age + "\nage:" + this.age)
  }
}
const fun2 = obj.fun1
fun2.apply(obj)
fun2.call(obj)
fun2.bind(obj)()

输出结果：

name:25
age:Tom
name:25
age:Tom
name:25
age:Tom

分析原因：

bind、call、apply更改了this的指向，因此输出了obj的属性

再次提出疑问，apply、call与bind的区别是什么

示例4

仔细观察我们在示例3中bind的用法，我们再次修改程序

var age = 26, name = "Jerry"
const obj = {
  age:25,
  name:"Tom",
  fun1(){
    console.log("name:" + this.age + "\nage:" + this.name)
  }
}
const fun2 = obj.fun1
fun2.apply(obj)
fun2.call(obj)
const fun3 = fun2.bind(obj)
fun3()

输出结果：

name:25
age:Tom
name:25
age:Tom
name:25
age:Tom

分析原因：

也就是说使用了apply和call则立刻就执行了该函数，而bind只是绑定了this对象，绑定后的函数内的this不再改变

明白了apply、call与bind之前的区别了，那apply与call的区别是什么呢？

示例5

我修改obj对象的函数，使其带有参数

var age = 26, name = "Jerry"
const obj = {
  age:25,
  name:"Tom",
  fun1(gender,address){
    console.log("name:" + this.age + "\nage:" + this.name)
    console.log("gender:" + gender + "\naddress:" + address)
  }
}
const fun2 = obj.fun1
fun2.apply(obj,["male","America"])
fun2.call(obj,"male","Canada")
fun2.bind(obj,"male","Africa")()

输出结果：

name:25
age:Tom
gender:male
address:America
name:25
age:Tom
gender:male
address:Canada
name:25
age:Tomgender:male
address:Africa

分析结果：

apply第二个参数可以是数组，而call和bind的参数则不必