++someVariable vs. someVariable++ in JavaScript

Question

In JavaScript you can use ++ operator before (pre-increment) or after the variable name (post-increment). What, if any, are the differences between these ways of incrementing a variable?

• Bergi: See also the language-agnostic Difference between i++ and ++i in a loop?
• palswim: I was thinking about this yesterday reading this response to the question about bad assumptions in C/C++. In all cases, can we guarantee that Javascript behaves this way? Or do you think it's bad practice to use the increment statement within a more complex statement at all?
• palswim: The previous comment is actually a copy of an answer (a non-answer, rather) I posted in 2010. I have deleted the answer, but Jon Skeet had replied with: "Looking at ECMA-262, it seems reasonably well-specified."
• shieldgenerator7: @palswim It's better to use them as standalone statements, and not within a more complex statement. Keeping it by itself improves readability and reduces confusion
• Yuval A.: This SO page makes things extremely unnecessarily complicated. i++ returns the value of i and increment it by one, ++i increment i by one and returns the incremented value. If i has 5, i++ will return 5 but then i will have 6, and ++i will return 6 and i will have 6.

Answer 1

Same as in other languages:

• ++x (pre-increment) means "increment the variable; the value of the expression is the final value"
• x++ (post-increment) means "remember the original value, then increment the variable; the value of the expression is the original value"

Now when used as a standalone statement, they mean the same thing:

x++;
++x;

The difference comes when you use the value of the expression elsewhere. For example:

x = 0;
y = array[x++]; // This will get array[0]
x = 0;
y = array[++x]; // This will get array[1]

• Chris: Curses, I nearly beat you to an answer had I not stopped to load up a practical jsfiddle answer. ;-)
• Keavon: What would this look like if you used + 1 instead of ++? Is there a way to increment before or after when adding numbers?
• Jean Jimenez: I would like to know why if you do this operation const r1 =(x++)+(x++); it does not produce the expected result according to your example.
• Jon Skeet: @JeanJimenez: Well it produces the result I expect. For example, if x starts off as 10, the value of r1 is 21, which is 10+11. The value of the first x++ expression is 10 and x is incremented to 11. The value of the second x++ expression is 11 and x is incremented to 12.
• Jean Jimenez: Dear @JonSkeet thanks for that super-fast response, I’m new to learning JavaScript and my confusion is regarding why one increments and the another doesn't.

Answer 2

• ++x increments the value, then evaluates and stores it.
• x++ evaluates the value, then increments and stores it.

var n = 0, m = 0;
alert(n++); /* Shows 0, then stores n = 1 */
alert(++m); /* Shows 1, then stores m = 1 */

Note that there are slight performance benefits to using ++x where possible, because you read the variable, modify it, then evaluate and store it. Versus the x++ operator where you read the value, evaluate it, modify it, then store it.

Answer 3

As I understand them if you use them standalone they do the same thing. If you try to output the result of them as an expression then they may differ. Try alert(i++) as compared to alert(++i) to see the difference. i++ evaluates to i before the addition and ++i does the addition before evaluating.

See http://jsfiddle.net/xaDC4/ for an example.

Answer 4

I've an explanation of understanding post-increment and pre-increment. So I'm putting it here.

Lets assign 0 to x

let x = 0;

Lets start with post-increment

console.log(x++); // Outputs 0

Why?

Lets break the x++ expression down

x = x;
x = x + 1;

First statement returns the value of x which is 0

And later when you use x variable anywhere, then the second statement is executed

Second statement returns the value of this x + 1 expression which is (0 + 1) = 1

Keep in mind the value of x at this state which is 1

Now lets start with pre-increment

console.log(++x); // Outputs 2

Why?

Lets break the ++x expression down

x = x + 1;
x = x;

First statement returns the value of this x + 1 expression which is (1 + 1) = 2

Second statement returns the value of x which is 2 so x = 2 thus it returns 2

Hope this would help you understand what post-increment and pre-increment are!

Answer 5

var a = 1;
var b = ++a;
alert('a:' + a + ';b:' + b); //a:2;b:2
var c = 1;
var d = c++;
alert('c:' + c + ';d:' + d); //c:2;d:1

jsfiddle

Answer 6

var x = 0, y = 0;
//post-increment: i++ returns value then adds one to it
console.log('x++ will log: ', x++); //0
console.log('x after x++ : ', x); //1
//pre-increment: adds one to the value, then returns it
console.log('++y will log: ', ++y); //1
console.log('y after ++y : ', y); //1

Answer 7

It is clearer and faster to use ++i if possible :

• ++i guarantees that you are using a value of i that will remains the same unless you change i
• i++ allows to use a value of i which will change in the "near future", it is not desirable if possible

Of course, it's not really much faster, only a little.