What is the difference if(“test”) and if(!!“test”)

Question

if("test")和if(!!"test")什么区别，只判断是假还是真；

Answer 1

The question has a double negation expressson, that converts the type to boolean.

eg

var x = "test";

x === true; // evaluates to false

var x = !!"test";

x === true; //evalutes to true

Answer 2

!! will convert a "truthy" value in true, and a "falsy" value on false.

"Falsy" values are the following:

• false
• 0 (zero)
• "" (empty string)
• null
• undefined
• NaN

If a variable x has any of these, then !!x will return false . Otherwise, !!x will return true .

On the practical side, there's no difference between doing if(x) and doing if(!!x) , at least not in javascript: both will enter/exit the if in the same cases.

EDIT: See http://www.sitepoint.com/blogs/2009/07/01/javascript-truthy-falsy/ for more info

Answer 3

!! does type conversion to a boolean, where you are just dropping it in to an if, it is AFAIK, pointless.

Answer 4

There is no functional difference. As others point out,

!!"test"

converts to string to a boolean.

Think of it like this:

!(!("test"))

First, "test" is evaluated as a string. Then !"test" is evaluated. As ! is the negation operator, it converts your string to a boolean. In many scripting languages, non-empty strings are evaluated as true, so ! changes it to a false. Then !(!"test") is evaluated, changing false to true.

But !! is generally not necessary in the if condition as like I mention it already does the conversion for you before checking the boolean value. That is, both of these lines:

if ("test")
if (!!"test")

are functionally equivalent.