Javascript: || operator for variable assignment

Question

I have two small examples code:

// example 1:
var x;
...
x = x || 'some value'; //assign some value to x if x has no value
...

// example 2:
var y;
...
y || (y = 'some value'); //assign some value to y if y has no value
...

My question is:
1. Will x be reassign it's value in example 1 when x already has value?
2. Is example 2 better then example 1?

(My english isn't good. Thanks for your reading and answering :D)

1) Depends on the value. 2) Better in which sense?

Felix Kling
– Felix Kling

2015-03-05 07:22:10 +00:00
Commented Mar 5, 2015 at 7:22 — Felix Kling
– Felix Kling, Commented Mar 5, 2015 at 7:22

Barmar · Accepted Answer · 2015-03-05 07:22:19Z

2

Yes. If x has a truthy value, it will be assigned back to itself. If it doesn't, the default 'some value' will be assigned to it.
There may be a tiny performance benefit in example 2, but example 1 is the standard idiom. Consistency with other programmers is useful because they'll understand your code more easily. Unless you're doing lots of default value initialization in a large loop, the performance gain should be negligible.

answered Mar 5, 2015 at 7:22

Barmar

789k57 gold badges555 silver badges669 bronze badges

Sign up to request clarification or add additional context in comments.

2 Comments

Minh Cường Over a year ago

thanks @Barmar, your answer is very clear and complete

Bergi Over a year ago

My guess would be that if (!y) y = 'some value'; is just as common (and definitely better / more widespread than example 2).

Dyrandz Famador · Accepted Answer · 2015-03-05 07:54:44Z

1

x will just reassign it's value since it has already a value

    var x = 'foo';
    x = x || 'some value';
  //  result: x = 'foo' //reassigned the value of x

In terms of readability, i would prefer the example1

edited Mar 5, 2015 at 7:54

answered Mar 5, 2015 at 7:28

Dyrandz Famador

4,5256 gold badges28 silver badges40 bronze badges

2 Comments

Felix Kling Over a year ago

The assignment x = x will still take place, so yes, x gets assigned the value it already has.

Dyrandz Famador Over a year ago

I got it now @Felix, kinda little bit confused a while ago.. :)

grodzi · Accepted Answer · 2015-03-05 07:22:04Z

-1

in example 1, javascript assigns at first pass value undefined to x. at second pass (when your code is running) x comes with value undefined at line

var x;

then || operator acts as following. Consider having :

left || right

if(left is not undefined (globally resolved to true) then return left else return right

so yes a value will be reassigned to x (either x itself or 1) As x resolved to undefined, or operator resolves right, and then assigns 1 to x. so x = x||1, ends in x's value === 1

second example no it adds uncomplicated logic (but still works)

answered Mar 5, 2015 at 7:22

grodzi

5,7111 gold badge17 silver badges16 bronze badges

2 Comments

Bergi Over a year ago

|| does not have a lot to do with undefined. And how does it matter that undefined is a global identifier?

grodzi Over a year ago

|| does have to do with undefined as undefined resolves to false for a boolean conversion. And || does check the boolean value of its operand. So in ( undefined || soomethingelse), the expression soomethingelse will be executed And I didn't mean undefined to be a global identifier. Globally refers to the left operand to be true (e.g false, null, 0..) so the right operand is not executed.

Collectives™ on Stack Overflow

Javascript: || operator for variable assignment

3 Answers 3

2 Comments

2 Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

2 Comments

2 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related