For a while, I have used "||" as the "or" indicator. One day, I was debugging some things in a console, and I accidentally put a single | instead of two. It still worked as expected.
console.log(0||1); // 1
console.log(0|1); // 1
Is there any difference? Here, there evidently isn't, but there might be some hidden difference that I don't know about. I apologize if this is a duplicate, but I assure you I have looked for the answer beforehand.
That is called a bitwise OR, meaning it ORs the individual bits that compose a value based on binary rules.
a b a OR b
0 0 0
0 1 1
1 0 1
1 1 1
For your example, 0 in binary is just 0000, and 1 in binary is 0001.
Thus 0|1 is:
0000 | 0001
Which, when we apply the table above between each binary digit of the two numbers:
0 or 0 = 0
0 or 0 = 0
0 or 0 = 0
0 or 1 = 1
give us 0001, which when converted to decimal becomes 1.
The way || (logical OR) behaves is using coercion rules which returns the first truthy item (or just the last item) in a sequence of ||.
Since 0 is falsy, 0 || 1, will return 1.
Just because the answers happen to be the same in these two situations, does not mean that the operations always produce equal results.
For instance:
2|3 === 3
2||3 === 2
| is a bitwise OR.
Performs the OR operation on each pair of bits. a OR b yields 1 if either a or b is 1
|| is a logical OR.
In your example, 0|1 will evaluate to 1, because ... 0000 | ... 0001 evaluates to ... 0001.
0||1 will evaluate to 1 because 0 is falsy so it'll return the right operand 1
