Result is getting twice by using console.log in javascript

Question

I am new in Javascript and play around with some stuff. In Chrome console I written this code.

var a;
console.log(a)

Output:undefined undefined

Can anybody pls explane me why I am getting twice?.

because what ever variable you put in console it will automatically displayed it's value — Beginner, Nov 15 '16 at 07:31
Because `console.log()` is a function, that returns `undefined` itself. — Itay, Nov 15 '16 at 07:32
`a` value is empty.so they print with `undefined`.You getting twice .i thing is the script in any loop.. — prasanth, Nov 15 '16 at 07:39
Isn't it just the Chrome debugger that return a cause for the undefined? — , Nov 15 '16 at 07:42

score 0 · Answer 1 · answered Nov 15 '16 at 07:31

0

console.log() outputs a value to the console because you called it and that is what console.log() does.

The return value of the last statement is then output to the console because you are running code on the console.

It is coincidence that those values are both undefined.

If you set var a = 1 then you would get 1 (what you logged) and then undefined (return value) instead.

answered Nov 15 '16 at 07:31

Quentin

`undefined` is the return value of the `console.log` function itself. – Itay Nov 15 '16 at 07:34
@Quentin So you mean one undefined for `consol.log()` and other which displaying `a` value in console as laty suggesting. – David Nov 15 '16 at 07:37

1 Answers1