Working with variables within javascript console.log function

Question

I am seeing sort of strange behaviour with javascript. I am new to this language, but from what I can see, if you increment a variable (or change it in any way) from within a console.log() method, this actually globally changes the variable.

var a = 0;

console.log(a); //prints 0

console.log(a++); //prints 0, a becomes 1
console.log(a++); //prints 1, a becomes 2
console.log(a++); //prints 2, a becomes 3

console.log(a); //prints 3

Is this something peculiar to javascript? I would have thought that the variable would not get affected globally and that the last print statement would show a as being 0.

Related: [javascript i++ vs ++i](https://stackoverflow.com/q/6867876/5894241) — Nisarg, Aug 07 '17 at 09:25
using `a++` is using a `post-increment` operation.. I don't see anything wrong here — kukkuz, Aug 07 '17 at 09:26
Yes that is perfectly normal. The console runs with the same scope as the `window`, or the page you are on. You're not running those bits of code in isolated scope, so they affect the page and its contents. — Reinstate Monica Cellio, Aug 07 '17 at 09:26
Its not the increment operator I have a problem with. It is the fact that statements from within a console.log() function can globally modify a variable. — Baba.S, Aug 07 '17 at 09:28
You're not using it "within" `console.log`, whatever that means. Your code is equivalent to `var b = a++; console.log(b);`. Not sure why you think inlining it into a function call would somehow isolate `a`. — deceze, Aug 07 '17 at 09:29

score 1 · Accepted Answer · answered Aug 07 '17 at 09:26

Using ++ will affect the variable, if you want to do it only for log purposes, you must use +1 , this is how it works in javascript ^^

var a = 0;

console.log(a); //prints 0

console.log(a+1);
console.log(a+1);
console.log(a+1);

console.log(a); //prints 0

Working with variables within javascript console.log function

1 Answers1