Question

Can anyone please explain how this line works.

 a & 3 || (b, c)

Does the first part translate to:

 a = a & 3;

or is that a ternary code and if true b is returned, else c? thanks

Answer 1

(Note on terminology: In 1 + 2, 1 and 2 are “operands” and + is an “operator”)

The first operator in that expression (&) is a bitwise ‘and’: https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Operators/Bitwise_Operators

The second operator (||) is a logical ‘or’, which returns the first operand if it’s truthy (if it would pass an if check), and the second one otherwise.

The third one (,) is the comma operator, which simply returns the second operand.

So, in plain English: Take the bitwise ‘and’ of a with 3 (which is 11 in binary), meaning that the resulting value will be one of 0, 1, 2, or 3 depending on the value of the first two bits in a. If it is not zero, return that value. Otherwise, return c, but b will also be evaluated.

For example, if a is 2, then a & 3 will be 10 & 11 == 10 (since 1 & 1 == 1 and 0 & 1 == 0), which is 2 and truthy (the only falsy number is 0 or 0.0), so that will be the return value of the whole expression, and b and c will not even be evaluated. On the other hand, if a is 4, then a & 3 will be 100 & 11 == 000, which is falsy, so (b, c) will be evaluated and the result will be that c is the return value.