Question: What's the difference between `|` and `||` in Javascript?

Furthermore, what's the difference between `&` and `&&`?

Answer: 
Another difference is that `||` uses shortcut evaluation. That is, it only evaluates the right side if the left side is `false` (or gets converted to false in a boolean context, e.g. `0`, `""`, `null`, etc.). Similarly, `&&` only evaluates the right side if the left side is `true` (or non-zero, non-empty string, an object, etc.). `|` and `&` always evaluate both sides because the result depends on the exact bits in each value.

The reasoning is that for `||`, if either side is true, the whole expression is true, so there's no need to evaluate any further. `&&` is the same but reversed.

The exact logic for `||` is that if the left hand side is "truthy", return that value (note that it is **not converted to a boolean**), otherwise, evaluate and return the right hand side. For `&&`, if the left hand side is "falsey", return it, otherwise evaluate and return the right hand side.

Here are some examples:

```
false && console.log("Nothing happens here");
true || console.log("Or here");
false || console.log("This DOES get logged");
"foo" && console.log("So does this");

if (obj && obj.property) // make sure obj is not null before accessing a property

```
Single | is bit wise OR operator . 
If you do 2 | 3 , it converts to binary and performs OR operation.
 01
 11
Results in 11 equal to 3.

Where as || operator checks if first argument is true, if it is true it returns else it goes to other operator.
2 || 3 returns 2 since 2 is true.