为什么类型判断用到Object.prototype.toString.call()进行类型判断，而不用typeof()呢？
然后翻了一下资料:
Typeof
在使用 typeof 运算符时采用引用类型存储值会出现一个问题，无论引用的是什么类型的对象，它都返回 "object"。
Object.prototype.toString.call():
ECMA 对Object.prototype.toString的解释
- Object.prototype.toString ( )
- When the toString method is called, the following steps are taken:
- If the this value is undefined, return "[object Undefined]".
- If the this value is null, return "[object Null]".
- Let O be the result of calling ToObject passing the this value as the argument.
- Let class be the value of the [[Class]] internal property of O.
- Return the String value that is the result of concatenating the three Strings "[object ", class, and "]".
然后举了个例子：
var oP = Object.prototype,
toString = oP.toString;
console.log(toString.call([123]));//[object Array]
console.log(toString.call('123'));//[object String]
console.log(toString.call({a: '123'}));//[object Object]
console.log(toString.call(/123/));//[object RegExp]
console.log(toString.call(123));//[object Number]
console.log(toString.call(undefined));//[object Undefined]
console.log(toString.call(null));//[object Null]
console.log(toString.call([123]));//[object Array]
console.log(toString.call('123'));//[object String]
console.log(toString.call({a: '123'}));//[object Object]
console.log(toString.call(/123/));//[object RegExp]
console.log(toString.call(123));//[object Number]
console.log(toString.call(undefined));//[object Undefined]
console.log(toString.call(null));//[object Null]