Simple substitute of assignment operators of logical ones in JavaScript?

Question

JavaScript has assignment operators corresponding to arithmetic ones: +=, -=, *=, /=, %=.

JavaScript also has assignment operators corresponding to bitwise ones: <<=, >>=, >>>=, &=, ^=, |=.

But it doesn't have assignment operators corresponding to logical ones: ||=, &&=.

Then, I can't do things like

aVeryLongVariableIdontWantToRepeat ||= 1;

In this other question it's explained why JS Java doesn't have such operators. I guess it's the same for JS.

But I want to know if there is a simple way to emulate them, avoiding

aVeryLongVariableIdontWantToRepeat = aVeryLongVariableIdontWantToRepeat || 1;

The other question you link to isn't about JavaScript, it's about Java. — T.J. Crowder, Dec 23 '13 at 22:35
@p.s.w.g I can't shorten things like `CanvasRenderingContext2D.prototype.clear` — Oriol, Dec 23 '13 at 22:38
@Oriol `var p = CanvasRenderingContext2D.prototype; p.clear = p.clear || ...`? — p.s.w.g, Dec 23 '13 at 22:40

score 2 · Accepted Answer · answered Dec 23 '13 at 22:34

2

No, there isn't. I feel like there should be more to this answer, but really, that's it. The shortest version of a = a || x is ... a = a || x.

answered Dec 23 '13 at 22:34

T.J. Crowder

There's also `a = a? a : x`, but that's a tad longer. – RobG Dec 23 '13 at 22:47

score 1 · Answer 2 · answered Dec 23 '13 at 23:15

1

It might help you to investigate writing your code using Coffeescript, which has the ||= operator available.

answered Dec 23 '13 at 23:15

anyaelise

score 0 · Answer 3 · answered Dec 23 '13 at 22:49

0

There isn't a shorter way: a = a || 1 is the simplest way to do it.

However, to avoid unnecessary assignment of values (slightly at the expense of readability) you can also do a || ( a = 1).

var a,b='x';
a || ( a = 1 );
b || ( b = 2 );
console.log( a + ', ' + b ); // Outputs "1, x"

answered Dec 23 '13 at 22:49

MT0

1

Of course, `a || (a = 1)` is no shorter than `a = a || 1` and we can count on the engine not to waste time on dead assignments. – T.J. Crowder Dec 23 '13 at 22:50

3 Answers3