为什么“typeof + ''”返回'number'？

Question

Let's try to type code below in console:

typeof + ''

This returns 'number', while typeof itself without argument throws an error. Why?

Answer 1

The unary plus operator invokes the internal ToNumber algorithm on the string. +'' === 0

typeof + ''
// The `typeof` operator has the same operator precedence than the unary plus operator,
// but these are evaluated from right to left so your expression is interpreted as:
typeof (+ '')
// which evaluates to:
typeof 0
"number"

Differently from parseInt , the internal ToNumber algorithm invoked by the + operator evaluates empty strings (as well as white-space only strings) to Number 0 . Scrolling down a bit from the ToNumber spec :

A StringNumericLiteral that is empty or contains only white space is converted to +0 .

Here's a quick check on the console:

>>> +''
<<< 0
>>> +' '
<<< 0
>>> +'\t\r\n'
<<< 0
//parseInt with any of the strings above will return NaN

For reference:

• Whats the significant use of Unary Plus and Minus operators? (SO)
• ES5 #9.3.1 ToNumber Applied to the String Type (ES5 Spec)
• Operator Precedence (MDN)

Answer 2

计算结果为typeof(+'') ，而不是(typeof) + ('')

Answer 3

Javascript interprets the following + '' as 0 so :

typeof + '' will echo `number'

To answer your second question, typeof takes an argument so if you call it by itself it will throw an error same thing if you call if by itself.