console.log(!status) in global scope producing unexpected result [duplicate]

Answer 1

See the below screenshot which says that status is a property in window. So this directly refers to the window.status. It is not advisable to reuse certain variables and properties that are having default meaning(reminding about keywords in oops languages)

Answer 2

Check the type for status like this:

typeof status

The output will be: "string".

Note that false and "false" are not same.

When you define:

var status = false; //which is actually "false"

That's why you are not getting true.

Answer 3

(function ( ) {
  var status = false;
  console.log(!status);
})();

This will give you expected output.

"status" is a predefined name of implementation-dependent JavaScript objects, methods, or properties. It is better avoid the identifiers as names of JavaScript variables. If you are using it outside the function then it means you want to implement any other thing by status value.

Answer 4

MDN

var statu = false;
console.log(!statu);

window.status exists already. so you can't use status variable .

Answer 5

window.status already exists (it is used to get/set the text of the browser's status bar) and when a value is assigned to it, it is converted to a string. If you do console.log( status ); you will see that status has the string value "false", which causes you to see the output false, since you effectively have !"false" and "false" is a truthy value in JavaScript.

If you do the same thing inside a function you'll get the expected output:

(function ( ) {
  var status = false;
  console.log(!status); // true
})();