Logical operator || returns number instead of boolean value

Question

I attached an example with two if conditions. The first if condition works as expected. The second if condition returns 11, but why? I know that the second if condition is wrong, but I would like to understand why Javascript returns in that case 11.

function exception(number) {
// if(number === 10 || number === 11) { // Working as expected
   if(number === 10 || 11) { // Why 11?
        console.log(number);
   }
}

function loop(f) {
    for (i = 0; i <= 100; i++) {
        f(i);
    }
}

loop(exception);

Possible duplicate of [Two pipe symbols (OR) in this Javascript line](http://stackoverflow.com/questions/10358823/two-pipe-symbols-or-in-this-javascript-line) — J0B, Jan 10 '16 at 14:15
Any non-zero number is considered truthy and will return `true` when converted to a boolean. — Nick Zuber, Jan 10 '16 at 14:19
fyi: sometimes it's handy to just do `if ( [10,11].indexOf(number) !== -1 )` — adeneo, Jan 10 '16 at 14:19
Possible duplicate of [What does the construct x = x || y mean?](http://stackoverflow.com/questions/2802055/what-does-the-construct-x-x-y-mean) — Michał Perłakowski, Jan 16 '16 at 02:58

score 6 · Answer 1 · edited May 23 '17 at 11:59

6

from this question.

(expr1 || expr2)

"Returns expr1 if it can be converted to true; otherwise, returns expr2."

source

So when expr1 is (or evaluates to) one of these 0,"",false,null,undefined,NaN, then expr2 is returned, otherwise expr1 is returned

edited May 23 '17 at 11:59

Community

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answered Jan 10 '16 at 14:16

J0B

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I like this answer because you actually address *why* the conditional expression is returning 11 – Nick Zuber Jan 10 '16 at 14:21
I think he knows that, the problem was that he thought `number === 10 || number === 11` would be the same as `number === 10 || 11` – CoderPi Jan 10 '16 at 14:26

CoderPi · Accepted Answer · 2016-01-10T14:18:59.467

5

Some Information about what you where trying to achieve:

number === 10 || number === 11 is the same as (number === 10) || (number === 11)
number === 10 || 11 is the same as (number === 10) || (11) it does not compare 11 to number here

Now let's have a closer look atnumber === 10 || 11 :

number === 10 will be true if number is of type number and equal to 10
if the first was false, it will evaluate the boolean value of the next statement: 11 (wich is accepted as true, for beeing a number not equal to 0)

edited Jan 10 '16 at 14:18

answered Jan 10 '16 at 14:14

CoderPi

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You can add explanation about the operator precedence – KarelG Jan 10 '16 at 14:16
@CodeiSir: Thanks. Now I got it. – HH. Jan 10 '16 at 15:54

score 0 · Answer 3 · answered Jan 10 '16 at 14:17

0

because Boolean(11) is true (try on your console)

so even if first condition is not true (if the number is not 10), then second condition will be true always

answered Jan 10 '16 at 14:17

gurvinder372

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Logical operator || returns number instead of boolean value

3 Answers3