Conditional assignment in JavaScript with Google Closure: | vs ||

Question

I am using Google Closure and LOVE conditional assignments. I regularly use the

var stuff = (condition) ? opt0: opt1;

method and have more recently been introduced to

var stuff = opt_param || defaultValue;

But I have noticed a strange behavior, since it seems the

var stuff = opt_param | defaultValue;

is aslso working (i.e. with the single |), but this appears to "type cast" the variable into a number. I have not been able to find any documentation on this, so here's my question:

What is the difference between the single | and the double || in terms of logic and assignment? And is my assumption regarding the number type cast correct, or just an arbitrary product of the special case(s) I've been fiddling with?

I don't see how this question is related to the closure tools. — Felix Kling, Apr 12 '13 at 09:44
possible duplicate of [what's the difference between ( | ) and ( || ) in javascript?](http://stackoverflow.com/questions/5690512/whats-the-difference-between-and-in-javascript) — Felix Kling, Apr 12 '13 at 09:45
You are absolutely right, Felix. I only thought it might, since the closure tools previously have been the only ones to tell me when I messed up in relation to "types"... Also; As far as I've been able to figure, it's not possible to get any results on special characters like the |-pipe, so that might be why I had missed the previously posted question. My bad. — NinjaTool, Apr 15 '13 at 10:51
No worries :) Since the introduction of a new search engine, you can search for special characters by putting them in quotes, e.g. `[javascript] "|" "||"`. — Felix Kling, Apr 15 '13 at 10:59

score 1 · Answer 1 · answered Apr 12 '13 at 09:24

1

| is bitwise OR:

3 | 4 = 7
0xFF00FF | 0x00FF00 = 0xFFFFFF

|| is logical OR, evaluates to the first argument if it is true, otherwise the second:

3 || 4 = 3
0 || 4 = 0

answered Apr 12 '13 at 09:24

Koterpillar

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VisioN · Accepted Answer · 2013-04-12T09:34:36.247

Briefly speaking, a single | refers to bitwise OR operator, which "performs the OR operation on each pair of bits. a OR b yields 1 if either a or b is 1". Example from MDN:

     9 (base 10) = 00000000000000000000000000001001 (base 2)
    14 (base 10) = 00000000000000000000000000001110 (base 2)
                   --------------------------------
14 | 9 (base 10) = 00000000000000000000000000001111 (base 2) = 15 (base 10)

While || is a logical OR that evaluates the first non-falsy statement in the assignment:

  123 || 0      === 123
false || "Text" === "Text"
   "" || true   === true

So basically these two operators are not interchangeable and may result in different values:

10 | 2  === 10
10 || 2 === 10

... however,

9 | 14  === 15
9 || 14 === 9

score 0 · Answer 3 · answered Apr 12 '13 at 09:28

The || operator is short-circuit OR. It assigns the first if it is true-like or the second. | on the other hand is Bit-wise operator. It performs a bitwise OR on the two parameters. If the first one evaluates to 0 result of bit-wise OR will be the second parameter.

It will give wrong results when both are numbers. Better go with ||.

Please have a look at Bitwise Operators on MDN.

Conditional assignment in JavaScript with Google Closure: | vs ||

3 Answers3